Proof By Induction Inequalities Questions, Answers and Solutions
a - answer s - solution v - video
Question 1
Prove by mathematical induction:
a) $2^n>2n$ for all $n>2$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
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To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ 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: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050040020
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) $n!>2^n$ for $n\geq4$
a
Answer and Solution are the same for proofs
To prove: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050040030
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050040040
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050040050
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050040060
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Answer by: ada
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
Prove by mathematical induction:
a) $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050045010
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 1
Prove by mathematical induction:
a) $2^n>2n$ for all $n>2$ and $n\in\mathbb{Z}^+$
b) $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
c) $n!>2^n$ for $n\geq4$
d) $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
e) $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
f) $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Question 2
Prove by mathematical induction:
a) $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ 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: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 1
Prove by mathematical induction:
a) $2^n>2n$ for all $n>2$ and $n\in\mathbb{Z}^+$
b) $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
c) $n!>2^n$ for $n\geq4$
d) $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
e) $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
f) $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Question 2
Prove by mathematical induction:
a) $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ 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: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Solutions
Question 1
a) Answer and Solution are the same for proofs
To prove: $2n<2^n$ for all $n>2$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=2\times3$
$=6$
$RHS=2^3$
$=8$
and $6<8$ so true
Step 2 - Assume true for $n=k$
$2k<2^k$
Step 3 - Show true for $n=k+1$
Need to show $2(k+1)<2^{k+1}$
So show $0<2^{k+1}-2(k+1)$
$RHS=2^{k+1}-2(k+1)$
$=2\times2^k-2(k+1)$
$>2\times2k-2k-2$ (from Step 2)
$=2k-2$
$>0$ (as $2k-2>0$ for $k>0$)
Which to shows that $2(k+1)<2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2k<2^k$, then $2(k+1)<2^{k+1}$. As $2n<2^n$ when $n=3$, it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $2^n<3^n$ for $n\geq1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$LHS=2^1$
$=2$
$RHS=3^1$
$=3$
and $2<3$ so true
Step 2 - Assume true for $n=k$
$2^k<3^k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}<3^{k+1}$
So show $0<3^{k+1}-2^{k+1}$
$RHS=3^{k+1}-2^{k+1}$
$=3\times3^k-2\times2^k$
$>3\times2^k-2\times2^k$ (from Step 2)
$=2^k$
$>0$ (as $2^k>0$ for $k\ge1$)
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k<3^k$, then $2^{k+1}<3^{k+1}$. As $2^n<3^n$ 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: $n!>2^n$ for $n\geq4$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=4$
$LHS=4!$
$=4\times3\times2\times1$
$=24$
$RHS=2^4$
$=16$
and $24>16$ so true
Step 2 - Assume true for $n=k$
$k!>2^k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)!>2^{k+1}$
So show $0<(k+1)!-2^{k+1}$
$RHS=(k+1)!-2^{k+1}$
$=(k+1)\times k!-2\times2^k$
$>(k+1)\times2^k-2\times2^k$ (from Step 2)
$=2^k(k+1-2)$
$=2^k(k-1)$
$>0$ (as $2^k>0$ and $(k-1)>0$ when for $k\ge4$
Which shows that $(k+1)!>2^{k+1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k!>2^k$, then $(k+1)!>2^{k+1}$. As $n!>2^n$ when $n=4$, it is also true for all $n\geq4$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $2^n>4n$ for $n\geq5$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=5$
$LHS=2^5=32$
$RHS=4\times5$
$=20$
and $32>20$ so true
Step 2 - Assume true for $n=k$
$2^k>4k$
Step 3 - Show true for $n=k+1$
Need to show $2^{k+1}>4(k+1)$
So show $0<2^{k+1}-4(k+1)$
$RHS=2^{k+1}-4(k+1)$
$=2\times2^k-4k-4$
$>2\times4k-4k-4$ (from Step 2)
$=4k-4$
$>0$ (as $4k-4>0$ for $k\ge5$)
Which shows that $2^{k+1}>4(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $2^k>4k$, then $2^{k+1}>4(k+1)$. As $2^n>4n$ when $n=5$, it is also true for all $n\geq5$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $n^2\ge2n$ for $n>1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=2$
$LHS=2^2=4$
$RHS=2\times2$
$=4$
and $4=4$ so true
Step 2 - Assume true for $n=k$
$k^2\geq2k$
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^2\geq2(k+1)$
So show $(k+1)^2-2(k+1)\geq0$
$LHS=(k+1)^2-2(k+1)$
$=k^2+2k+1-2k-2$
$=k^2-1$
$\geq2k-1$ (from Step 2)
$\geq0$ (as $2k-1\geq0$ for $k>1$)
Which shows that $(k+1)^2\geq2(k+1)$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2\geq2k$, then $(k+1)^2\geq2(k+1)$. As $n^2\geq2n$ when $n=2$, it is also true for all $n>1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $n^2<4^{n-1}$ for $n\geq3$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=3$
$LHS=3^2=9$
$RHS=4^{3-1}=4^2$
$=16$
and $9<16$ so true
Step 2 - Assume true for $n=k$
$k^2<4^{k-1}$
Step 3 - Show true for $n=k+1$
Need to show ${(k+1)}^2<4^{(k+1)-1}$
So show $0<4^{(k+1)-1}-{(k+1)}^2$
$RHS=4^{(k+1)-1}-{(k+1)}^2$
$=4^k-k^2-2k-1$
$=4\times4^{k-1}-k^2-2k-1$
$>4\times k^2-k^2-2k-1$ (from Step 2)
$=3k^2-2k-1$
$=(3k+1)(k-1)$
$>0$ (as $(3k+1)>0$ and $(k-1)>0$ for $k\geq3$)
Which shows that ${(k+1)}^2<4^{(k+1)-1}$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^2<4^{k-1}$, then ${(k+1)}^2<4^{(k+1)-1}$. As $n^2<4^{n-1}$ when $n=3$, it is also true for all $n\ge3$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt n}<2\sqrt n-1$ for $n\ge2$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=2$
$LHS=1+\frac{1}{\sqrt2}$
$=\frac{\sqrt2+1}{\sqrt2}$
$=\frac{2+\sqrt2}{2}$
$\approx1.71$ ($2$dp)
$RHS=2\sqrt2-1$
$\approx1.83$ ($2$dp)
$>LHS$
So true for $n=2$
Step 2 - Assume true for $n=k$
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}<2\sqrt k-1$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}<2\sqrt{k+1}-1$
So show $[1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}+\frac{1}{\sqrt{k+1}}]-[2\sqrt{k+1}-1]<0$ $LHS=1+\frac{1}{\sqrt2}+\frac{1}{\sqrt3}\dotsb+\frac{1}{\sqrt k}]+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$<2\sqrt k-1+\frac{1}{\sqrt{k+1}}-2\sqrt{k+1}+1$
$=2\sqrt k-2\sqrt{k+1}+\frac{1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2\sqrt{k+1}\sqrt{k+1}+1}{\sqrt{k+1}}$
$=\frac{2\sqrt k\sqrt{k+1}-2(k+1)+1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-2k-1}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k+1)}{\sqrt{k+1}}$
$<\frac{2\sqrt{k^2+k}-k(2k+2)}{\sqrt{k+1}}$
$=\frac{2\sqrt{k^2+k}-(2k^2+2k)}{\sqrt{k+1}}$
$=\frac{2(\sqrt{k^2+k}-(k^2+k))}{\sqrt{k+1}}$
$=\frac{2\left(\sqrt{k^2+k}-\sqrt{\left(k^2+k\right)^2}\right)}{\sqrt{k+1}}$
$<0$ (as $\sqrt{\left(k^2+k\right)^2}>\sqrt{k^2+k}$ for $k>0$)
which was what was required to prove
Step 4 - Conclusion
If true for $n=k$ then true for $n=k+1$. As true for $n=2$ it is also true for all $n\ge2$ and $n\in\mathbb{Z}^+$ by mathematical induction.