Question

求1³＋2³+……＋n³ 公式

如果三次方换成其他整数有没有相应的公式 即1的a次方一直加到n的a次方
谢谢

Answer 1

1³+2³+3³+······+n³
=（1+2+3+……+n)²  

 

证明：

Answer 2

1³+2³+……+n³=[n(n+1)/2]²

公式推导：
利用立方差公式：
(n+1)^4-n^4=[(n+1)^2+n^2][(n+1)^2-n^2]
=(2n^2+2n+1)(2n+1)
=4n^3+6n^2+4n+1

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

各式相加有
(n+1)^4-1=4*(1^3+2^3+3^3...+n^3)+6*(1^2+2^2+...+n^2)+4*(1+2+3+...+n)+n

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

http://zhidao.baidu.com/link?url=Tjp_VSuL3oseFuJPAj2q0lGfhVsJjmYgbwW4TWAnnWHiE5jKfRjhug1UfWB216oWhe2aHrYTwZYoewr4bqO3LcQ2dmIy-tFAQhxRVorQTiW

Answer 3

n^3
= n(n+1)(n+2) - 3n^2-2n
= n(n+1)(n+2) - 3n(n+1) +n
= (1/4)[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)] - [ n(n+1)(n+2) -(n-1)n(n+1)] + (1/2)[n(n+1)-(n-1)n]
1^3+2^3+...+n^3
= (1/4)n(n+1)(n+2)(n+3) - n(n+1)(n+2) + (1/2)n(n+1)
=(1/4)n(n+1)[(n+2)(n+3)-4(n+2)+2]
=(1/4)n(n+1)(n^2+n)
= [(1/2)n(n+1)]^2

Answer 4

引用jssqnju的回答：
1³+2³+3³+······+n³=（1+2+3+……+n)²

证明：

左右相加后应该是f(n)-f(0)＝1³+2³+…+n³