數歸問題 , help help !!

let the statement
P(n): 4^n + 5^n is divisible by 9 for all odd positive integers n
For n = 1
4+5=9 is divisible by 9
∴ the statment is true for n = 1.

Assume the statment is true for n = k (where k is odd)
i.e. 4^k + 5^k is divisible by 9
let 4^k + 5^k=9M

For n = k + 2,
4^(k + 2) + 5^(k + 2)
=16(4^k)+25(5^k)
=16(4^k + 5^k)+9(5^k)
=16(9M)+9(5^k)
=9(16M+5^k)
which is divisible by 9
∴ the statment is true for n = k + 1.

By the principle of mathematical induction, the statment is true for all positive odd integers n


2007-01-31 22:49:53 補充：
自己最後打錯了添﹐因為急住幫你∴ the statment is true for n = k + 2.

Let P(n) be the proposition that
4^n+5^n is divisible by 9
where n is an odd number

When n = 1
4 + 5 = 9 which is divisible by 9
∴P(1) is true.

Assume P(K) is true, where K is some odd number.
i.e. 4^K + 5^K = 9N, where N is an integer
Consider P(K+2)
4^(K+2) + 5^(K+2) = 16∙4^K +25∙5^K
= 16(9N-5^K) +25∙5^K
= 16(9N)+(25-16) 5^K
= 9(16N+5^K) which is divisible by 9
∴P(K+2) is true.
∴By M.I., P(n) is true for all odd numbers n.

You should check the question. It should be valid on odd numbers only
For 4^2 + 5^2 = 41, 4^4 + 5^4 = 881, ....(all even indices) are not divisible by 9