mathematical induction 5

When n=1,3^2+2^(2-2)=9+1=10,which is divisible by 5.
Assume n=k, 3^2k+2^(2k-2)=5M,where M is a integer.
when n=k+1
=3^[2(k + 1)] + 2^[2(k + 1) - 2]
=3^(2k + 2) + 2^(2k+2 - 2)
=(3^2k)(3^2) + 2^(2k - 2)(2^2)
=(3^2k)(9) + [ 2^(2k - 2) ](4)
=(3^2k)(4 + 5) + [2^(2k - 2)](4)
= [(3^2k)(4) + (3^2k)(5)] + [2^(2k - 2)](4)
=(3^2k)(4) + (3^2k)(5) + [ 2^(2k - 2) ](4)
Take out the common factor,4
= (4)[(3^2k) + 2^(2k-2)] + (3^2k)(5)
= (4)(5M)+ (3^2k)(5)
Take out the common factor ,5
=(5)(4M+3^2k)
so,n=k+1 is divisible by 5
By the principal of mathematical induction,P(k+1) is true of all positive integer

Your question should state like this
3^2n + 2^(2n-2) is divisible by 5 for all natural numbers n.
Do you need to prove?




2006-11-16 19:47:04 補充：
The first responser was misleded because 3^2n+2^2n-2 is watched as3^2n+(2^2n)-2 so he (she) proved it is wrong.If you want me to prove, I shall prove it later.

Let P(n) be the proposition 3^(2n)+2^(2n)-2 is divisible by 5
When n=1
3^(2n)+2^(2n)-2
=3^2+2^2-2
=11
Therefore S(1) is false.
Therefore S(n) is not true for all natural number n.
The proposition was disproved.