Re: number....

From: José Carlos Santos (jcsantos_at_fc.up.pt)
Date: 10/03/04 

Date: Sun, 03 Oct 2004 17:37:01 +0100

mina_world wrote:

> show that
> 2^(2n + 1) = {9*(n^2)} - (3n) + 2 (mod 54)
> for all positive integer n.

Put f(n) = 2^(2n + 1) - 9n^2 + 3n - 2. You want to prove that 54
divides each f(n). Since this is clearly true when n = 1 (because
f(1) = 0), it's enough to prove that f(n + 1) - f(n) is always a
multiple of 54. But

    f(n + 1) - f(n) = 6(4^n - 1 - 3n).

So, you have to prove that 4^n - 1 - 3n is always a multiple of 9,
which is quite easy (remember that 4 = 3 + 1).

Best regards,

Jose Carlos Santos

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