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