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Define a sequence ($a_i$) i∈ Natural Numbers, recursively by $a_1 = 3, a_2 = −6$, and, for all $ n ≥ 2,\; a_{n+1} = a_n + 2a_{n−1} + 3.$ Prove $3$|$a_n$ for all $n ∈ \mathbb N$.

I have tried this problem, but I can't get past the inductive step, where I need to prove $2a_{n-1}$ is divisible by $3$. Is there a way to finish the proof?

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    $\begingroup$ Do you mean $$a_{n+1}=a_n+2a_{n-1}+3$$? $\endgroup$ – Dr. Sonnhard Graubner Feb 24 at 19:42
  • $\begingroup$ Hi, welcome. Please don't put the question in the title but in the content of the question :) $\endgroup$ – Stan Tendijck Feb 24 at 19:57
  • $\begingroup$ and please use MathJax $\endgroup$ – J. W. Tanner Feb 24 at 20:02
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We can prove $3|a_i$ and $3|a_{i+1}$ by induction.

Clearly $3|a_1=3$ and $3|a_2=-6$.

If $3|a_{n-1}$ and $3|a_n,$ then, because a linear combination of integers divisible by $3$ is divisible by $3,$

$3|a_{n+1}=a_n+2a_{n-1}+3$ and $3|a_n$.

Note that we didn't have to calculate $a_n.$

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  • $\begingroup$ How can we assume that $a_(n-1)$ is divisible by 3? $\endgroup$ – beepbeepboop123123 Feb 25 at 0:10
  • $\begingroup$ Our base case has two consecutive terms divisible by 3. For the inductive step we assume two consecutive terms are divisible by 3 and show the next pair is. Since the later term of the previous pair is the earlier term of the next pair, it’s given that it’s a multiple of 3 $\endgroup$ – J. W. Tanner Feb 25 at 1:03

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