是否存在常数A,B使等式:1(N^2-1^2)+2(N^2-2^2)+3(N^2-3^2)+……+N(N^2-N^2)=

记住常用求和公式
1+2+3+...+n=n(n+1)/2
1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6
1^3+2^3+3^3+...+n^3=[n(n+1)/2]^2

因1(N^2-1^2)+2(N^2-2^2)+3(N^2-3^2)+…+N(N^2-N^2)
=N^2(1+2+3+...+N)-(1^3+2^3+3^3+...+N^3)
=N^2*N(N+1)/2-[N(N+1)/2]^2
=[N^2(N+1)(N-1)]/4
则A=1,B=-1或A=-1,B=1

当N=1时,1(N^2-1^2)=1(1^2-1^2)=0,N^2(N+1)(N-1)]/4=1^2(1+1)(1-1)]/4=0,等式成立
假设N=k时有1(k^2-1^2)+2(k^2-2^2)+3(k^2-3^2)+…+k(k^2-k^2)=[k^2(k+1)(k-1)]/4
则当N=k+1时,
1[(k+1)^2-1^2]+2[(k+1)^2-2^2]+3[(k+1)^2-3^2]+...+k[(k+1)^2-k^2]+(k+1)[(k+1)^2-(k+1)^2]
=[1(k^2-1^2)+2(k^2-2^2)+3(k^2-3^2)+…+k(k^2-k^2)] + (2k+1)(1+2+3+...+k) （注意到(k+1)[(k+1)^2-(k+1)^2]=0）
=[k^2(k+1)(k-1)]/4+[(2k+1)(k+1)k]/2={k(k+1)[k(k-1)+2(2k+1)]}/4=[(k+1)^2(k+2)k]/4
因此1(N^2-1^2)+2(N^2-2^2)+3(N^2-3^2)+…+N(N^2-N^2)=[N^2(N+1)(N-1)]/4对一切N属于N*都成立