S1 = 1 = 1^4
S1+S3 = 16 = 2^4
S1+S3+S5 = 81 = 3^4
S1+S3+S5+...+S(2n-1) = n^4
當(dāng)n=1時(shí),成立
設(shè)當(dāng)n=k(>1)時(shí) S1+S3+S5+...+S(2k-1) = k^4
當(dāng)n= k+1時(shí) 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
得證