S1 = 1 = 1^4
S1+S3 = 16 = 2^4
S1+S3+S5 = 81 = 3^4
S1+S3+S5+...+S(2n-1) = n^4
当n=1时,成立
设当n=k(>1)时 S1+S3+S5+...+S(2k-1) = k^4
当n= k+1时 S1+S3+S5+...+S(2k-1)+S(2k+1) = k^4 + S(2k+1)
= k^4 + (1+2k)*(2k)/2 *(2k+1) + (1+2k+1)/2 * (2k+1)
= k^4 + 4k^3 + 6k^2 + 4k + 1
= (k+1)^4
得证