Mathematical Induction 怎樣做啊!

1(a)題目應該係32n+1 + 2n+2，而非32n+1 + 22n+2
let S(n) = 32n+1 + 2n+2
S(1) = 33 + 23 = 35 = 7(5)
S(2) = 35 + 24= 259 = 7(37)
∴p should be 7

(b)Let S(n) be the proposition:
‘32n+1 + 2n+2 is divisible by 7’
When n = 1,
32n+1 + 2n+2 = 35 = 7(5)
∴S(1) is true.
Assume that S(k) is true,
i.e. 32k+1 + 2k+2 = 7m , where m is an integer
When n = k+1,
32(k+1)+1 + 2(k+1)+2 = 32k+1×(32﹣1)+2k+2×(2﹣1)

32k+3+2k+3﹣7m
= 8×32k+1+2k+2
= 7×32k+1+32k+1+2k+2
= 7×32k+1+7m

32k+3+2k+3
= 7×32k+1+14m
= 7(32k+1+2m)
∴S(k+1) is also true.
By the principle of Mathematical induction,S(n) is true for all positive integers n.

do you mean 3^(2n+1)+2^(2n+2) or 3^2n+1 + 2^2n+2?