f.4m.i. come help

n = 1 P(1) = 3^4 - 8 - 9 = 64
n = 2 P(2) = 3^6 - 16 - 9 = 704 = 64 x 11
n = 3 P(3) = 3^8 - 24 - 9 = 6528 = 64 x 102.
For n = k, P(k) = 3^(2k + 2) - 8k - 9 = 64M
so 3^(2k + 2) = 64M + 8k + 9
For n = k + 1
P(k + 1) = 3^[2(k + 1) + 2] - 8(k + 1) - 9
= 3^(2k + 4) - 8k - 8 - 9
= 3^(2k + 2)(3^2) - 8k - 17
= 9(64M + 8k + 9) - 8k - 17
= 576M + 72k + 81 - 8k - 17
= 576M + 64k - 64
= 64(9M + k - 1)
so 3^(2n + 2) - 8n - 9 is divisible by 64.

2010-12-14 22:35:11 補充：
Correction : The last 3 lines should be ' = 576M + 64k + 64' and '= 64(9M + k + 1)'.......

when n=1, 3^2n+2 -8n -9 = -6
when n=2, 3^2n+2 -8n -9 = 58
when n=3, 3^2n+2 -8n -9 = 698

since when n=1,2,3, 3^2n+2 -8n -9 is not divisible by 64
it cant be proved by M.I.

btw, 你係米出錯題?