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Suppose that $a_0, a_1, a_2, \dots$ is a sequence defined as follows. $$a_0 = 2, a_1 = 4, a_2 = 6 \text{, and } a_k = 7 a_{k-3} \text{ for all integers $k \ge 3$.}$$ Prove that $a_n$ is even for all integers $n \ge 0$.

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  • $\begingroup$ What are your thoughts on the problem? $\endgroup$ – kingW3 Oct 20 '15 at 21:49
  • $\begingroup$ Well I have no idea how to prove it, I know how to prove the base case and write the inductive step but not applying inductive hypothesis and proving it's even! $\endgroup$ – isharailanga Oct 20 '15 at 21:55
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Except the basic case is trivial and assuming that the statement holds for each $k \leq n - 1$ you have $a_n = 7a_{n-3} $. By hypothesis $a_{k-3} = 2\gamma$, $\gamma$ integer, so $a_n = 14 \gamma$ which implies that 2 divides $a_n $.

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  • $\begingroup$ what about 'k+1'th term of a?How can we prove it's also divisible!Or Do I really have to prove that too :( $\endgroup$ – isharailanga Oct 20 '15 at 22:36
  • $\begingroup$ I made a mistake, the induction is basic case and assuming that it holds up to n -1 then we prove the case n. Now since it is true up to n you could verify n + 1... but you don't need to do that because of the induction you just proved. About the divisibility... an integer is even if it is the product of 2 times another integer, and such property assuming it holds up to n -1 also hold to n as we proved. $\endgroup$ – user8469759 Oct 20 '15 at 23:23

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