the question goes as follows: Use proof by induction to show that $2^{n+1} + 5 \times 9^n$ is divisible by $7$.
I did a few things like $f(k+1)-f(k)$ to get to $2^{k+1} + 5 \times 9^k \times 8$ but I dont understand what to do next
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How about $f(k+1) - 2f(k)$? – user317176 Nov 23 '21 at 18:43
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@DougM i tried that and got to -5 x 9^(k+1) did i do something wrong? – Coder Cody Nov 23 '21 at 18:48
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Write $9=7+2$ and use the binomial theorem. – Mark Viola Nov 23 '21 at 18:58
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@MarkViola thank you i slowly understood it, i never thought about it that way ty – Coder Cody Nov 23 '21 at 19:00
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Let $P(n)$ be the statement that $2^{n+1}+5\times 9^n$ is divisible by $7$.
Base case: $n=0$, trivial.
Inductive Step: assume the statement is true for some $n = k\in\mathbb N$, we want to show the statement for $n=k+1$.
Then
\begin{align*} 2^{k+2} + 5\times 9^{k+1} &= 2\times 2^{k+1} + 5\times 9^k\times 9\\ &= 2\times (2^{k+1}+5\times 9^k) + (5\times 9^k)\times 7\\&=\cdots \end{align*}
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um how did you factor out 2 from 5 x 9^k and where did the 7 +(5 x9^k) come from? sorry – Coder Cody Nov 23 '21 at 18:53
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