1的立方加2的立方加到n的立方

答案是[n(n+1)/2]^2。解题步骤如下：

n^4-(n-1)^4

=[n^2-(n-1)^2][n^2+(n-1)^2]

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

=4n^3-6n^2+4n-1

2^4-1^4=4*2^3-6*2^2+4*2-1

3^4-2^4=4*3^3-6*3^2+4*3-1

4^4-3^4=4*4^3-6*4^2+4*4-1

......

n^4-(n-1)^4=4n^3-6n^2+4n-1

各等式全部相加

n^4-1^4=4*(2^3+3^3+...+n^3)-6*(2^2+3^2+...+n^2)+4(2+3+4+...+n)-(n-1)

n^4-1^4=4*(1^3+2^3+3^3+...+n^3)-6*(1^2+2^2+3^2+...+n^2)+4(1+2+3+4+...+n)-(n-1)-2

n^4-1=4*(1^3+2^3+3^3+...+n^3)-6*n(n+1)(2n+1)/6+4*n(n+1)/2-n-1

n^4-1=4*(1^3+2^3+3^3+...+n^3)-n(n+1)(2n+1)+2n(n+1)-n-1

n^4-1=4*(1^3+2^3+3^3+...+n^3)-n(n+1)(2n+1)+2n(n+1)-n-1

4*(1^3+2^3+3^3+...+n^3)

=n^4-1+n(n+1)(2n+1)-2n(n+1)+n+1

=n^4-1+(n+1)(2n^2-n)+n+1

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

=n^4+2n^3+n^2

=(n^2+n)^2

=(n(n+1))^2

1^3+2^3+3^3+...+n^3

=[n(n+1)/2]^2

扩展资料

等差数列{an}的通项公式为：an=a1+(n-1)d。前n项和公式为：Sn=n*a1+n(n-1)d/2或Sn=n(a1+an)/2。

等比数列前n项之和Sn=A1(1-q^n)/(1-q)=A1(q^n-1)/(q-1)=(A1q^n)/(q-1)-A1/(q-1)。