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cyb3r · #3 November 16th, 2008, 09:49 AM

I was trying to do it the past 2 hours, and got a solution, but I'm not very convinced about it.

Here is what I did :

1) We have H(0) = 3 which is divisable by 3
2) We have H(1) = -3 which is divisable by 3, too
3) Let's have $H(2)$ which is the equivalance relation $2H(n-1)-H(n-2)+6$

$2H(n-1) = H(1)$ - divisable by 3
$H(n-2) = H(0)$ - divisable by 3
$6 = 6$ - divisable by 3

So H(2) is divisable by 3 too, and since we have a recurrence relation, this applies to all positive integers following the relation.