是否存在常数a,b,c,使一切都有1^3+2^3+3^3+……+n^3=1/4n^2(an^2+bn+c)

(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;
所以递推得：
n^4-(n-1)^4=4(n-1)^3+6(n-1)^2+4(n-1)+1;
...
2^4-1=4*1^3+6*1^2+4*1+1;
所有等式左右相加得：
(n+1)^4-1=4(1^3+2^3+...n^3)+6(1^2+2^2+...+n^2)+4(1+2+...+n)+n;
(n+1)^4-1=4(1^3+2^3+...n^3)+6*[n*n+1)(2n+1)/6]+4[(n+1)n/2]+n;
(n+1)^4-1=4(1^3+2^3+...n^3)+n*(n+1)(2n+1)+2n(n+1)+n;
4(1^3+2^3+...n^3)=(n+1)^4-[n*(n+1)(2n+1)+2n(n+1)+n+1];
=n^4+2n3+n^2=n^2(n^2+2n+1);
1^3+2^3+...n^3=1/4n^2(n^2+2n+1);
存在常数a,b,c,使一切都有1^3+2^3+3^3+……+n^3=1/4n^2(an^2+bn+c)
a=1;b=2,c=1