等差級數和之公式的證明

You statement is wrong, it should be 1^3+2^3+...n^3=[n(n+1)/2]^2.
Proof:
(r+1)^4-r^4=r^4+4r^3+6r^2+4r+1-r^4
=4r^3+6r^2+4r+1
take summation on both side from r=1 to n
(n+1)^4-1=4(1^3+2^3+...n^3)+6[n(n+1)(2n+1)/6]+4n(n+1)/2+n
4(1^3+2^3+...n^3)=(n+1)^4-1-n(n+1)(2n+1)-2n(n+1)-n
=n^4+4n^3+6n^2+4n+1-1-2n^3-3n^2-n-2n^2-2n-n
=n^4+2n^3+n^2
=n^2(n^2+2n+1)
=[n(n+1)]^2
1^3+2^3+...n^3=[n(n+1)/2]^2

2010-10-30 10:01:38 補充：
'n^3' means n(立方)
Also,the other method is mathematical induction

可能無人能 證明：1(立方)+2(立方)+3(立方)+4(立方)+......+n(立方)=[n(n+1)/2]立方，應該是=[n(n+1)/2]平方才對吧？