Proof By Induction Series Questions, Answers and Solutions
a - answer s - solution v - video
Question 1
Prove by mathematical induction for $n\ge1$ and $n\in\mathbb{Z}^+$:
a) $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
a
Answer and Solution are the same for proofs
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050010010
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Question by: ada
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Answer by: ada
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) $1+3+5+7+\dotsb+(2n-1)=n^2$
a
Answer and Solution are the same for proofs
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050010020
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Question by: ada
Answer by: ada
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Question by: ada
Answer by: ada
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
a
Answer and Solution are the same for proofs
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050010030
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
a
Answer and Solution are the same for proofs
To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050010040
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To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
a
Answer and Solution are the same for proofs
To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050010050
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To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
a
Answer and Solution are the same for proofs
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050010060
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Question by: ada
Answer by: ada
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
a
Answer and Solution are the same for proofs
To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
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To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
h) $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$
a
Answer and Solution are the same for proofs
To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050010080
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To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 1
Prove by mathematical induction for $n\ge1$ and $n\in\mathbb{Z}^+$:
a) $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
b) $1+3+5+7+\dotsb+(2n-1)=n^2$
c) $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
d) $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
e) $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
f) $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
g) $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
h) $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) Answer and Solution are the same for proofs
To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
h) Answer and Solution are the same for proofs
To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 1
Prove by mathematical induction for $n\ge1$ and $n\in\mathbb{Z}^+$:
a) $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
b) $1+3+5+7+\dotsb+(2n-1)=n^2$
c) $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
d) $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
e) $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
f) $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
g) $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
h) $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) Answer and Solution are the same for proofs
To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
h) Answer and Solution are the same for proofs
To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Solutions
Question 1
a) Answer and Solution are the same for proofs
To prove: $1+2+3+4+\dotsb+n=\frac{n(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(1+1)}{2}$
$=\frac{1\times2}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
$LHS=1+2+3+4+\dotsb+k+(k+1)$
$=[ \,1+2+3+4+\dotsb+k]\,+(k+1)$
$=[ \,\frac{k(k+1)}{2}]\,+(k+1)$ (using Step 2)
$=\frac{k(k+1)}{2}+(k+1)$
$=\frac{k(k+1)}{2}+\frac{2(k+1)}{2}$
$=\frac{k(k+1)+2(k+1)}{2}$
$=\frac{k^2+k+2k+2}{2}$
$=\frac{k^2+3k+2}{2}$
$=\frac{(k+1)(k+2)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+2+3+4+\dotsb+k=\frac{k(k+1)}{2}$, then $1+2+3+4+\dotsb+k+(k+1)=\frac{(k+1)(k+2)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $1+3+5+7+\dotsb+(2n-1)=n^2$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=1^2$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+3+5+7+\dotsb+(2k-1)=k^2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
$LHS=1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)$
$=[ \,1+3+5+7+\dotsb+(2k-1)]\,+(2(k+1)-1)$
$=[ \,k^2]\,+(2(k+1)-1)$ (using Step 2)
$=k^2+2k+2-1$
$=k^2+2k+1$
$=(k+1)^2$
$=RHS$
Step 4 - Conclusion
If $1+3+5+7+\dotsb+(2k-1)=k^2$, then $1+3+5+7+\dotsb+(2k-1)+(2(k+1)-1)=(k+1)^2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $1+4+7+10+\dotsb+(3n-2)=\frac{n(3n-1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=1$
$RHS=\frac{1(3\times1-1)}{2}$
$=\frac22$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
$LHS=1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)$
$=[ \,1+4+7+10+\dotsb+(3k-2)]\,+(3(k+1)-2)$
$=[ \,\frac{k(3k-1)}{2}]\,+(3(k+1)-2)$ (using Step 2)
$=\frac{k(3k-1)}{2}+3k+3-2$
$=\frac{k(3k-1)}{2}+3k+1$
$=\frac{k(3k-1)}{2}+\frac{2(3k+1)}{2}$
$=\frac{k(3k-1)}{2}+\frac{6k+2}{2}$
$=\frac{k(3k-1)+6k+2}{2}$
$=\frac{3k^2-k+6k+2}{2}$
$=\frac{3k^2+5k+2}{2}$
$=\frac{(k+1)(3k+2)}{2}$
$=\frac{(k+1)(3(k+1)-1)}{2}$
$=RHS$
Step 4 - Conclusion
If $1+4+7+10+\dotsb+(3k-2)=\frac{k(3k-1)}{2}$, then $1+4+7+10+\dotsb+(3k-2)+(3(k+1)-2)=\frac{(k+1)(3(k+1)-1)}{2}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $1^2+2^2+3^2+4^2+\dotsb+n^2=\frac{n(n+1)(2n+1)}{6}$
Step 1 - Prove true for $n=1$
$LHS=1^2$
$=1$
$RHS=\frac{1(1+1)(2\times1+1)}{6}$
$=\frac{1(2)(3)}{6}$
$=\frac66$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$LHS=1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2$
$=[ \,1^2+2^2+3^2+4^2+\dotsb+k^2]\,+(k+1)^2$
$=[ \,\frac{k(k+1)(2k+1)}{6}]\,+(k+1)^2$ (using Step 2)
$=\frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$
$=\frac{k(k+1)(2k+1)+6(k+1)^2}{6}$
$=\frac{(k+1)(k(2k+1)+6(k+1))}{6}$ (by factorising numerator)
$=\frac{(k+1)(2k^2+k+6k+6)}{6}$
$=\frac{(k+1)(2k^2+7k+6)}{6}$
$=\frac{(k+1)(k+2)(2k+3)}{6}$
$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
$=RHS$
Step 4 - Conclusion
If $1^2+2^2+3^2+4^2+\dotsb+k^2=\frac{k(k+1)(2k+1)}{6}$, then $1^2+2^2+3^2+4^2+\dotsb+k^2+(k+1)^2=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $2^1+2^2+2^3+2^4+\dotsb+2^n=2^{n+1}-2$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=2^{1+1}-2$
$=2^2-2$
$=4-2$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
$LHS=2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}$
$=[ \,2^1+2^2+2^3+2^4+\dotsb+2^k]\,+2^{k+1}$
$=[ \,2^{k+1}-2]\,+2^{k+1}$
$=2^{k+1}+2^{k+1}-2$
$=2\times2^{k+1}-2$
$=2^1\times2^{k+1}-2$
$=2^{(k+1)+1}-2$
$=RHS$
Step 4 - Conclusion
If $2^1+2^2+2^3+2^4+\dotsb+2^k=2^{k+1}-2$, then $2^1+2^2+2^3+2^4+\dotsb+2^k+2^{k+1}=2^{(k+1)+1}-2$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{n(n+1)}=\frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{1\times2}$
$=\frac{1}{2}$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}=\frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}=\frac{k+1}{(k+1)+1}$
$LHS=\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{1}{1\times2}+\frac{1}{2\times3}+\dotsb+\frac{1}{k(k+1)}\right]+\frac{1}{(k+1)(k+2)}$
$=\left[\frac{k}{k+1}\right]+\frac{1}{(k+1)(k+2)}$ (using Step 2)
$=\frac{k(k+2)}{(k+1)(k+2)}+\frac{1}{(k+1)(k+2)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)(k+1)}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) Answer and Solution are the same for proofs
To prove: $\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times n^2-1}=\frac{n}{2n+1}$
Step 1 - Prove true for $n=1$
$LHS=\frac{1}{4\times1^2-1}$
$=\frac{1}{3}$
$RHS=\frac{1}{2\times1+1}$
$=\frac13$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}=\frac{k}{2k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}=\frac{k+1}{2(k+1)+1}$
$LHS=\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{1}{4\times1^2-1}+\frac{1}{4\times2^2-1}+\dotsb+\frac{1}{4\times k^2-1}\right]+\frac{1}{4\times(k+1)^2-1}$
$=\left[\frac{k}{2k+1}\right]+\frac{1}{4\times(k+1)^2-1}$ (using Step 2)
$=\frac{k}{2k+1}+\frac{1}{4k^2+8k+3}$
$=\frac{k}{2k+1}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{k(2k+3)}{(2k+1)(2k+3)}+\frac{1}{(2k+1)(2k+3)}$
$=\frac{2k^2+3k+1}{(2k+1)(2k+3)}$
$=\frac{(2k+1)(k+1)}{(2k+1)(2k+3)}$
$=\frac{k+1}{2k+3}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then true for $n=k+1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
h) Answer and Solution are the same for proofs
To prove: $1^3+2^3+3^3+\dots+n^3=(1+2+3+\dots+n)^2$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=1^3$
$=1$
$RHS=1^2$
$=1$
So true for $n=1$
Step 2 - Assume true for $n=k$
$1^3+2^3+3^3+\dots+k^3=(1+2+3+\dots+k)^2$
Step 3 - Show true for $n=k+1$
Need to show $1^3+2^3+3^3+\dots+k^3+(k+1)^3=(1+2+3+\dots+k+(k+1))^2$
$LHS=1^3+2^3+3^3+\dots+k^3+(k+1)^3$
$=(1+2+3+\dots+k)^2+(k+1)^3$ (using Step 2)
as $1+2+\dots+k=\frac12k(k+1)$ (sum of arithmetic series)
then $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$
so $=\frac14k^2(k+1)^2+(k+1)^3$
$=(k+1)^2\left(\frac14k^2+(k+1)\right)$
$=\frac14(k+1)^2(k^2+4k+4))$
$=\frac14(k+1)^2(k+2)^2$
$=(1+2+\dots+k+(k+1))^2$ (using $(1+2+\dots+k)^2=\frac14k^2(k+1)^2$ but with $k+1$ in place of $k$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If As it is rue for $n=1$ and true for $n=k+1$, assuming it is true for $n=k$ then it is true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.