Proof By Induction Division Questions, Answers and Solutions
a - answer s - solution v - video
Question 1
Prove by mathematical induction:
a) $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
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To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050030020
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To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) $5^n-1$ is divisible by $4$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
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Question by: ada
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To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050030040
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050030050
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050030060
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Question by: ada
Answer by: ada
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
Prove by mathematical induction:
a) $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.
v
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.
Question ID: 10050060020
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Question by: ada
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Question by: ada
Answer by: ada
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.
Question 1
Prove by mathematical induction:
a) $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
b) $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
c) $5^n-1$ is divisible by $4$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
d) $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
e) $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
f) $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Question 2
Prove by mathematical induction:
a) $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.
Question 1
Prove by mathematical induction:
a) $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
b) $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
c) $5^n-1$ is divisible by $4$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
d) $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
e) $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
f) $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Question 2
Prove by mathematical induction:
a) $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.
Solutions
Question 1
a) Answer and Solution are the same for proofs
To prove: $6^n+4$ is divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$6^0+4=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$6^k+4=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
$6^{k+1}+4=6\times6^k+4$
(using $6^k+4=5m\to6^k=5m-4$)
$=6(5m-4)+4$
$=30m-24+4$
$=30m-20$
$=5(6m-4)$
and $5(6m-4)$ is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $6^k+4$ is divisible by $5$, then $6^{k+1}+4$ is divisible by $5$. As $6^n+4$ is divisible by $5$ when $n=0$, it is also divisible by $5$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $8^n-1$ is divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-1=7$
$7$ is divisible by $7$
Step 2 - Assume true for $n=k$
$8^k-1=7m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-1$ is divisible by $7$
$8^{k+1}-1$
$=8\times8^k-1$
$=8\times(7m+1)-1$ (using Step 2: $8^k-1=7m\to8^k=7m+1$)
$=56m+8-1$
$=56m+7$
$=7(8m+1)$
and $7(8m+1)$ is divisible by $7$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-1$ is divisible by $7$, then $8^{k+1}-1$ is divisible by $7$. As $8^n-1$ is divisible by $7$ when $n=1$, it is also divisible by $7$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $5^n-1$ is divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$5^1-1=4$
$4$ is divisible by $4$
Step 2 - Assume true for $n=k$
$5^k-1=4m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}-1$ is divisible by $4$
$5^{k+1}-1$
$=5\times5^k-1$
$=5\times(4m+1)-1$ (using Step 2: $5^k-1=4m\to5^k=4m+1$)
$=20m+5-1$
$=20m+4$
$=4(5m+1)$
and $4(5m+1)$ is divisible by $4$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k-1$ is divisible by $4$, then $5^{k+1}-1$ is divisible by $4$. As $5^n-1$ is divisible by $4$ when $n=1$, it is also divisible by $4$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $n^3-7n+9$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$0^3-7\times0+9=9$
$9$ is divisible by $3$
Step 2 - Assume true for $n=k$
$k^3-7k+9=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $(k+1)^3-7(k+1)+9$ is divisible by $3$
$(k+1)^3-7(k+1)+9$
$=k^3+3k^2+3k+1-7k-7+9$
$=(k^3-7k+9)+3k^2+3k+1-7$
$=(3m)+3k^2+3k-6$ (using Step 2: $k^3-7k+9=3m$)
$=3m+3k^2+3k-6$
$=3(m+k^2+k-2)$
and $3(m+k^2+k-2)$ is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $k^3-7k+9$ is divisible by $3$, then $(k+1)^3-7(k+1)+9$ is divisible by $3$. As $n^3-7n+9$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
e) Answer and Solution are the same for proofs
To prove: $8^n-3^n$ is divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$
$8^1-3^1=5$
$5$ is divisible by $5$
Step 2 - Assume true for $n=k$
$8^k-3^k=5m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $8^{k+1}-3^{k+1}$ is divisible by $5$
$8^{k+1}-3^{k+1}$
$=8\times8^k-3^{k+1}$
$=8(5m+3^k)-3^{k+1}$ (rearranging Step 2 to get $8^k=5m+3^k$)
$=(40m+8\times3^k)-3\times3^k$
$=40m+8\times3^k-3\times3^k$
$=40m+5\times3^k$
$=5(8m+3^k)$, which is divisible by $5$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $8^k-3^k$ is divisible by $5$, then $8^{k+1}-3^{k+1}$ is divisible by $5$. As $8^n-3^n$ is divisible by $5$ when $n=1$, it is also divisible by $5$ for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
f) Answer and Solution are the same for proofs
To prove: $5^n+2\times11^n$ is divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=0$
$5^0+2\times11^0=1+2\times1$
$=3$
$3$ is divisible by $3$
Step 2 - Assume true for $n=k$
$5^k+2\times11^k=3m$, where $m$ is an integer
Step 3 - Show true for $n=k+1$
Need to show $5^{k+1}+2\times11^{k+1}$ is divisible by $3$
$5^{k+1}+2\times11^{k+1}$
$=5\times(5^k)+2\times11^{k+1}$
$=5(3m-2\times11^k)+2\times11^{k+1}$ (by rearranging Step 2 to get $5^k=3m-2\times11^k$)
$=(15m-10\times11^k)+2\times11\times11^k$
$=15m-10\times11^k+22\times11^k$
$=15m+12\times11^k$
$=3(5m+4\times11^k)$, which is divisible by $3$
Therefore it is true for $n=k+1$
Step 4 - Conclusion
If $5^k+2\times11^k$ is divisible by $3$, then $5^{k+1}+2\times11^{k+1}$ is divisible by $3$. As $5^n+2\times11^n$ is divisible by $3$ when $n=0$, it is also divisible by $3$ for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $x^n-y^n$ can be divided by $x-y$ for $n\geq1$, $x\ne y$ and $n\in\mathbb{Z}^+$
Step 1 - Prove true for $n=1$ and $n=2$
When $n=1$
$x^1-y^1$ is divisible by $x-y$
When $n=2$
$x^2-y^2=(x-y)(x+y)$
So divisible by $x-y$
Step 2 - Assume true for $n=k$ and $n=k+1$
So, $x^k-y^k$ is divisible by $x-y$
There must be some polynomial $P(x,y)$ (with variables $x$ and $y$) that when multiplied by $x-y$ is $x^k-y^k$
$x^k-y^k=(x-y)P(x,y)$
Similarly $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$
Step 3 - Show true for $n=k+2$
Need to show when $x^{k+2}-y^{k+2}=(x-y)R(x,y)$ for some polynomial $R(x,y)$
$x^{k+2}-y^{k+2}=x\times x^{k+1}-y\times y^{k+1}$
$=[x\times x^{k+1}-x\times y^{k+1}]+[y\times x^{k+1}-y\times y^{k+1}]-(y\times x^{k+1}-x\times y^{k+1})$
$=x[x^{k+1}-y^{k+1}]+y[x^{k+1}-y^{k+1}]-(y\times x\times x^k-x\times y\times y^k)$
$=(x+y)[x^{k+1}-y^{k+1}]-xy(x^k-y^k)$
$=(x+y)(x-y)Q(x,y)-xy(x-y)P(x,y)$ (from Step 2)
$=(x-y)[(x+y)Q(x,y)-xyP(x,y)]$
$=(x-y)R(x,y)$ (let $R(x,y)=(x+y)Q(x,y)-xyP(x,y))$
Therefore has been shown
Step 4 - Conclusion
If $x^k-y^k$ and $x^{k+1}-y^{k+1}$ are divisible by $x-y$ then we have proven that $x^{k+2}-y^{k+2}$ is also divisible by $x-y$. As true when $n=1$ and $n=2$ it is also true for all positive integers by mathematical induction.