大家都知道1+2+3+……+n=n(n+1)/2

1^2+2^2+3^2+……+n^2=n(n+1)(2n+1)/6,

1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2,現在咱們推導

1^4+2^4+3^4+……+n^4:

我們知道(n-1)^5=n^5-5n^4+10n^3-10n^2+5n-1

那麼有

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

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

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

……

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

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

1^5-0^5=5*1^4-10*1^3+10*1^2-5*1+1

左邊相加等於右邊,

左邊之和為n^5,

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

令1^4+2^4+3^4+……+n^4=M

得到n^5=5M-10[n(n+1)/2]^2+10n(n+1)(2n+1)/6-5n(n+1)/2+n,

上面的式子中有個M,解出

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

即1^4+2^4+3^4+……+n^4=n(n+1)(2n+1)(3n^2+3n-1)/30.

補:

1^5+2^5+3^5+……+n^5

=n^2(n+1)^2(2n^2+2n-1)/12