Proof By Induction Sigma Notation Questions, Answers and Solutions
a - answer s - solution v - video
Question 1
Prove by mathematical induction for $n\in\mathbb{Z}^+$:
a) $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
a
Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
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: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050020010
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
a
Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
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: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050020020
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Question by: ada
Answer by: ada
Something wrong? Copy Question ID and contact us
Submit yours! Copy Question ID and click here
Question by: ada
Answer by: ada
To prove: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
a
Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
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: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050020030
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To prove: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) $\sum_{x=1}^n x\times x! = (n+1)!-1$
a
Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
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: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050020035
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Answer by: ada
To prove: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
a
Answer and Solution are the same for proofs
To prove: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050020040
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Question by: ada
Answer by: ada
To prove: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
a
Answer and Solution are the same for proofs
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050020050
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
a
Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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: 10050020060
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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 1
Prove by mathematical induction for $n\in\mathbb{Z}^+$:
a) $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
b) $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
c) $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
d) $\sum_{x=1}^n x\times x! = (n+1)!-1$
e) $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
f) $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
g) $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
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: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
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: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
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: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
As true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
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: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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 1
Prove by mathematical induction for $n\in\mathbb{Z}^+$:
a) $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
b) $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
c) $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
d) $\sum_{x=1}^n x\times x! = (n+1)!-1$
e) $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
f) $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
g) $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
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: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
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: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
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: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
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: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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.
Solutions
Question 1
a) Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n x^3 = \frac{n^2{(n+1)}^2}{4}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x^3$
$=1^3$
$=1$
$RHS=\frac{1^2{(1+1)}^2}{4}$
$=\frac{1\times2^2}{4}$
$=\frac{1\times4}{4}$
$=\frac44$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
$LHS=\sum_{x=1}^{k+1} x^3$
$=1^3+2^3+3^3+\dotsb+k^3+(k+1)^3$
$=[\,1^3+2^3+3^3+\dotsb+k^3]\,+(k+1)^3$
$=[\,\sum_{x=1}^k x^3]\,+(k+1)^3$
$=[\,\frac{k^2{(k+1)}^2}{4}]\,+(k+1)^3$ (using Step 2)
$=\frac{k^2{(k+1)}^2}{4}+\frac{4(k+1)^3}{4}$
$=\frac{k^2{(k+1)}^2+4(k+1)^3}{4}$
$=\frac{(k+1)^2(k^2+4(k+1))}{4}$ (by factorising)
$=\frac{(k+1)^2(k^2+4k+4)}{4}$
$=\frac{(k+1)^2(k+2)^2}{4}$
$=\frac{(k+1)^2((k+1)+1)^2}{4}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x^3 = \frac{k^2{(k+1)}^2}{4}$, then $\sum_{x=1}^{k+1} x^3=\frac{(k+1)^2{((k+1)+1)}^2}{4}$
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: $\sum_{x=1}^n x(x+1) = \frac{n(n+1)(n+2)}{3}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x(x+1)$
$=1(1+1)$
$=1\times2$
$=2$
$RHS=\frac{1(1+1)(1+2)}{3}$
$=\frac{1(2)(3)}{3}$
$=\frac{6}{3}$
$=2$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$LHS=\sum_{x=1}^{k+1} x(x+1)$
$=1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)+(k+1)((k+1)+1)$
$=[\,1(1+1)+2(2+1)+3(3+1)+\dotsb+k(k+1)]\,+(k+1)((k+1)+1)$
$=[\,\sum_{x=1}^{k} x(x+1)]\,+(k+1)((k+1)+1)$
$=[\,\frac{k(k+1)(k+2)}{3}]\,+(k+1)((k+1)+1)$ (using Step 2)
$=\frac{k(k+1)(k+2)}{3}+\frac{3(k+1)((k+1)+1)}{3}$
$=\frac{k(k+1)(k+2)+3(k+1)((k+1)+1)}{3}$
$=\frac{(k+1)(k(k+2)+3((k+1)+1))}{3}$ (by factorising)
$=\frac{(k+1)(k^2+2k+3(k+2))}{3}$
$=\frac{(k+1)(k^2+2k+3k+6)}{3}$
$=\frac{(k+1)(k^2+5k+6)}{3}$
$=\frac{(k+1)(k+2)(k+3)}{3}$
$=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x(x+1)=\frac{k(k+1)(k+2)}{3}$, then $\sum_{x=1}^{k+1} x(x+1)=\frac{(k+1)((k+1)+1)((k+1)+2)}{3}$
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: $\sum_{x=1}^n \frac{1}{x(x+1)} = \frac{n}{n+1}$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1\frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}$
$=\frac{1}{1\times2}$
$=\frac12$
$RHS=\frac{1}{1+1}$
$=\frac12$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
$LHS=\sum_{x=1}^{k+1} \frac{1}{x(x+1)}$
$=\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{1}{1(1+1)}+\frac{1}{2(2+1)}+\frac{1}{3(3+1)}+\dotsb+\frac{1}{k(k+1)}}\right]+\frac{1}{(k+1)((k+1)+1)}$
$=\left[{\frac{k}{k+1}}\right]+\frac{1}{(k+1)((k+1)+1)}$ (using Step 2)
$=\frac{k}{k+1}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k((k+1)+1)}{(k+1)((k+1)+1)}+\frac{1}{(k+1)((k+1)+1)}$
$=\frac{k(k+2)+1}{(k+1)((k+1)+1)}$
$=\frac{k^2+2k+1}{(k+1)(k+2)}$
$=\frac{(k+1)^2}{(k+1)(k+2)}$
$=\frac{k+1}{k+2}$
$=\frac{k+1}{(k+1)+1}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k \frac{1}{x(x+1)} = \frac{k}{k+1}$, then $\sum_{x=1}^{k+1} \frac{1}{x(x+1)} = \frac{k+1}{(k+1)+1}$
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: $\sum_{x=1}^n x\times x! = (n+1)!-1$
Step 1 - Prove true for $n=1$
$LHS=\sum_{x=1}^1 x\times x!$
$=1\times1!$
$=1$
$RHS=(1+1)!-1$
$=2!-1$
$=2\times1-1$
$=2-1$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k x\times x! = (k+1)!-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
$LHS=\sum_{x=1}^{k+1} x\times x!$
$=\sum_{x=1}^{k} x\times x!+(k+1)\times(k+1)!$
$=(k+1)!-1+(k+1)\times(k+1)!$ (using Step 2)
$=(k+1)!(k+1+1)-1$ (by factorising)
$=(k+1)!(k+2)-1$
$=(k+2)!-1$
$=((k+1)+1)!-1$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=1}^k x\times x! = (k+1)!-1$, then $\sum_{x=1}^{k+1} x\times x! = ((k+1)+1)!-1$
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: $\sum_{x=0}^n r^x = \frac{1-r^{n+1}}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 r^x $
$=r^0$
$=1$
$RHS=\frac{1-r^{0+1}}{1-r}$
$=\frac{1-r^1}{1-r}$
$=\frac{1-r}{1-r}$
$=1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
$LHS=\sum_{x=0}^{k+1} r^x$
$=r^0+r^1+r^2+\dotsb+r^k+r^{k+1}$
$=[\,r^0+r^1+r^2+\dotsb+r^k]\,+r^{k+1}$
$=[\,\sum_{x=0}^k r^x]\,+r^{k+1}$
$=[\,\frac{1-r^{k+1}}{1-r}]\,+r^{k+1}$ (using Step 2)
$=\frac{1-r^{k+1}}{1-r}+\frac{(1-r)(r^{k+1})}{1-r}$
$=\frac{(1-r^{k+1})+(1-r)(r^{k+1})}{1-r}$
$=\frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$
$=\frac{1-r^{k+2}}{1-r}$
$=\frac{1-r^{(k+1)+1}}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k r^x = \frac{1-r^{k+1}}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{1-r^{(k+1)+1}}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $\sum_{x=0}^n ar^x = \frac{a(1-r^{n+1})}{1-r}$, $r\neq1$
Step 1 - Prove true for $n=0$
$LHS=\sum_{x=0}^0 ar^x$
$=ar^0$
$=a$
$RHS=\frac{a(1-r^{0+1})}{1-r}$
$=\frac{a(1-r^1)}{1-r}$
$=\frac{a(1-r)}{1-r}$
$=a$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=0}^{k+1} ar^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
$LHS=\sum_{x=0}^{k+1} ar^x$
$=ar^0+ar^1+ar^2+\dotsb+ar^k+ar^{k+1}$
$=[\,ar^0+ar^1+ar^2+\dotsb+ar^k]\,+ar^{k+1}$
$=[\,\sum_{x=0}^k ar^x]\,+ar^{k+1}$
$=[\,\frac{a(1-r^{k+1})}{1-r}]\,+ar^{k+1}$ (using Step 2)
$=\frac{a(1-r^{k+1})}{1-r}+\frac{(1-r)(ar^{k+1})}{1-r}$
$=\frac{a(1-r^{k+1})+(1-r)(ar^{k+1})}{1-r}$
$=\frac{a-ar^{k+1}+ar^{k+1}-ar^{k+2}}{1-r}$
$=\frac{a-ar^{k+2}}{1-r}$
$=\frac{a-ar^{(k+1)+1}}{1-r}$
$=\frac{a(1-r^{(k+1)+1})}{1-r}$
$=RHS$
Step 4 - Conclusion
If $\sum_{x=0}^k ar^x = \frac{a(1-r^{k+1})}{1-r}$, then $\sum_{x=0}^{k+1} r^x = \frac{a(1-r^{(k+1)+1})}{1-r}$
As true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
g) Answer and Solution are the same for proofs
To prove: $\sum_{x=1}^n (-1)^x\times x^2 = \frac{(-1)^nn(n+1)}{2}$
Step 1 - Prove true for $n=1$
$LHS=(-1)^1\times1^2$
$=-1$
$RHS=\frac{(-1)^1\times1\times(1+1)}{2}$
$=\frac{-2}{2}$
$=-1$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sum_{x=1}^k (-1)^x\times x^2 = \frac{(-1)^kk(k+1)}{2}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sum_{x=1}^{k+1} (-1)^x\times x^2=\frac{(-1)^{k+1}(k+1)((k+1)+1)}{2}$
$LHS=\sum_{x=1}^{k+1} (-1)^x\times x^2$
$=\left[\sum_{x=1}^{k}(-1)^x\times x^2\right]+(-1)^{k+1}\times(k+1)^2$
$=\left[\frac{(-1)^kk(k+1)}{2}\right]+(-1)^{k+1}\times(k+1)^2$ (using Step 2)
$=\frac{(-1)^kk(k+1)}{2}+\frac{2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^kk(k+1)+2(-1)^{k+1}\times(k+1)^2}{2}$
$=\frac{(-1)^k(k+1)[k+2\times(-1)\times(k+1)]}{2}$
$=\frac{(-1)^k(k+1)[k-2k-2]}{2}$
$=\frac{(-1)^k(k+1)[-k-2]}{2}$
$=\frac{(-1)^{k+1}(k+1)[k+2]}{2}$
$=RHS$
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.