Math Help Forum - View Single Post

clic-clac · #4 November 16th, 2008, 10:02 AM

Well what we want to prove is that for every $n \in \mathbb{N}, 3|H(n)$ .
We know it's true for $H(0)$ and $H(1)$ . Now, let $k\geq 2$ be an integer, and assume that $H(k-2)$ and $H(k-1)$ are divisible by $3$ .
We want to prove that $3$ also divides $H(k)$ .

$H(k)=2H(k-1) + H(k-2) - 6$

So the question is: Is a sum of integers divisible by $3$ divisible by $3$ ?

If yes, the your proof is complete!

The following users thank clic-clac for this useful post:
cyb3r
$Donate to MHF$