# 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 $n\in\mathbb{Z}^+$

$6^0+4=5$

$5$ is divisible by $5$

$6^k+4=5m$, where $m$ is an integer

$6^{k+1}+4=6\times6^k+4$

(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 $n=k+1$

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 $n\in\mathbb{Z}^+$

$6^0+4=5$

$5$ is divisible by $5$

$6^k+4=5m$, where $m$ is an integer

$6^{k+1}+4=6\times6^k+4$

(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 $n=k+1$

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 $n\in\mathbb{Z}^+$

$8^1-1=7$

$7$ is divisible by $7$

$8^k-1=7m$, where $m$ is an integer

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 $n=k+1$

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 $n\in\mathbb{Z}^+$

$8^1-1=7$

$7$ is divisible by $7$

$8^k-1=7m$, where $m$ is an integer

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 $n=k+1$

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.

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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 $n\in\mathbb{Z}^+$

$5^1-1=4$

$4$ is divisible by $4$

$5^k-1=4m$, where $m$ is an integer

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 $n=k+1$

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 $n\in\mathbb{Z}^+$

$5^1-1=4$

$4$ is divisible by $4$

$5^k-1=4m$, where $m$ is an integer

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 $n=k+1$

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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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 $n\in\mathbb{Z}^+$

$0^3-7\times0+9=9$

$9$ is divisible by $3$

$k^3-7k+9=3m$, where $m$ is an integer

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 $n=k+1$

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 $n\in\mathbb{Z}^+$

$0^3-7\times0+9=9$

$9$ is divisible by $3$

$k^3-7k+9=3m$, where $m$ is an integer

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 $n=k+1$

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.

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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 $n\in\mathbb{Z}^+$

$8^1-3^1=5$

$5$ is divisible by $5$

$8^k-3^k=5m$, where $m$ is an integer

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 $n=k+1$

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 $n\in\mathbb{Z}^+$

$8^1-3^1=5$

$5$ is divisible by $5$

$8^k-3^k=5m$, where $m$ is an integer

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 $n=k+1$

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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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 $n\in\mathbb{Z}^+$

$5^0+2\times11^0=1+2\times1$

$=3$

$3$ is divisible by $3$

$5^k+2\times11^k=3m$, where $m$ is an integer

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 $n=k+1$

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 $n\in\mathbb{Z}^+$

$5^0+2\times11^0=1+2\times1$

$=3$

$3$ is divisible by $3$

$5^k+2\times11^k=3m$, where $m$ is an integer

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 $n=k+1$

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.

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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 $n\in\mathbb{Z}^+$

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 $x-y$

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 $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$

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

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 $n\in\mathbb{Z}^+$

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 $x-y$

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 $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$

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

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

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 $n\in\mathbb{Z}^+$

$6^0+4=5$

$5$ is divisible by $5$

$6^k+4=5m$, where $m$ is an integer

$6^{k+1}+4=6\times6^k+4$

(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 $n=k+1$

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__- ConclusionIf $6^k+4$ 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 $n\in\mathbb{Z}^+$

$8^1-1=7$

$7$ is divisible by $7$

$8^k-1=7m$, where $m$ is an integer

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 $n=k+1$

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__- ConclusionIf $8^k-1$ 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 $n\in\mathbb{Z}^+$

$5^1-1=4$

$4$ is divisible by $4$

$5^k-1=4m$, where $m$ is an integer

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 $n=k+1$

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__- ConclusionIf $5^k-1$ 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 $n\in\mathbb{Z}^+$

$0^3-7\times0+9=9$

$9$ is divisible by $3$

$k^3-7k+9=3m$, where $m$ is an integer

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 $n=k+1$

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__- ConclusionIf $k^3-7k+9$ 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 $n\in\mathbb{Z}^+$

$8^1-3^1=5$

$5$ is divisible by $5$

$8^k-3^k=5m$, where $m$ is an integer

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 $n=k+1$

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__- ConclusionIf $8^k-3^k$ 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 $n\in\mathbb{Z}^+$

$5^0+2\times11^0=1+2\times1$

$=3$

$3$ is divisible by $3$

$5^k+2\times11^k=3m$, where $m$ is an integer

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 $n=k+1$

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__- ConclusionIf $5^k+2\times11^k$ 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 $n\in\mathbb{Z}^+$

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 $x-y$

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 $x^{k+1}-y^{k+1}=(x-y)Q(x,y)$

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

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__- ConclusionIf $x^k-y^k$ 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.