1^4+2^4+3^4+4^4+5^4+```+n^4=? 并用非反证法证明下，谢啦

(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

http://baike.baidu.com/view/1722042.htm

∑i^4＝∑(i+1)i(i-1)i+∑i
=∑(i+1)i(i-1)(i-2)+∑i^2+2∑(i+1)i(i-1)
=∑(i+1)i(i-1)(i-2)+∑i(i-1)+∑i+2∑(i+1)i(i-1)
=24∑C(4,i+1)+12∑C(3,i+1)+∑C(1,i)+2∑C(2,i)
=24C(5,n+2)+12C(4,n+2)+C(2,n+1)+2C(3,n+1)