Question

如何求四次方和五次方的各项和,公式+推导

Answer 1

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

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

……

2^5-1^5=5*1^4+10*1^3+10*1^2+5*1+1

全加起来

(n+1)^5-1^5=5*(1^4+2^4+3^4+4^4+……+n^4)+10*(1^3+2^3+3^3+4^3+……+n^3)+10*(1^2+2^2+3^2+4^4+……+n^2)+5*(1+2+3+4+……+n)+n

因为1^3+2^3+3^3+4^3+……+n^3=[n(n+1)/2]^2

1^2+2^2+3^2+4^4+……+n^2=n(n+1)(2n+1)/6

1+2+3+4+……+n=n(n+1)/2

所以1^4+2^4+3^4+4^4+……+n^4

={[(n+1)^5-1^5]-10*[n(n+1)/2]^2-10*n(n+1)(2n+1)/6-5*n(n+1)/2-n}/5

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