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
對比an^4+bn^2+c
a=1/4,b=-1/4,c=0
所以存在常數(shù)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對一切正整數(shù)n都成立.
補充：
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