1^4+2^4+3^4+...+n^4=? 

試解:

(k+1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1         (1)

對(1)，從k = 1 到 k = n 求和，得到

(n+1)^5 = (n+1) + 5(1^4+2^4+...+n^4) + 10(1^3+2^3+...+n^3) + 

        10(1^2+2^2+...+n^2) + 5(1+2+...+n)

5S = (n+1)^5 - (n+1) - 5n^2(n+1)^2/2 - 5n(n+1)(2n+1)/3 - 5n(n+1)/2 

5S = (n+1)[(n+1)^4 - 1 - 5n^2(n+1)/2 - 5n(2n+1)/3 - 5n/2] 

整理後得

5S = n(n+1)[6n^3 + 9n^2 + n - 1]/6 

5S = n(n+1)(2n+1)[3n^2 + 3n - 1]/6 

S = n(n+1)(2n+1)(3n^2+3n-1)/30