是否存在常数abc,使得等式1*2^2+2*3^2+.+n(n+1)^n=n(n+1)(an^2+bn+c)/12成立?

1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2) =(1+2+..+n)*n^2-(1^3+2^3+..+n^3) 其中：1+2+3+..+n=n*(n+1)/2 1^3+2^3+...+n^3=[n(n+1)/2]^2 所以： 1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2) =(1+2+..+n)*n^2-(1^3+2^3+..+n^3) ＝n^3*(n+1)/2 -[n(n+1)/2]^2 =n*(n+1)(2n^2-n^2-n)/4 =(n^2+n)(n^2-n)/4 =(n^4-n^2)/4 对比an^4+bn^2+c a=1/4,b=-1/4,c=0 所以存在常数a、b、c,使等式1*(n^2-1^2)+2*(n^2-2^2)...+n(n^2-n^2)=an^4+bn^2+c对一切正整数n都成立. 补充： 1^3+2^3+3^3+……+n^3=[n(n+1)/2]^2 (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 1^3+2^3+...+n^3=[n(n+1)/2]^2