(x+1)(x+2)(x+3)(x+4)^(x+n)展开式中x的(n－2)次幂的系数怎么求?

Question

(x+1)(x+2)(x+3)(x+4)^(x+n)展开式中x的(n－2)次幂的系数怎么求?

Answer 1

=(1×2+1×3+…+1×n)+(2×3+…+2×n)+… +(n-1)n
=1×2+(1+2)×3+…+[1+…+(n-1)]n
则上式每项的通式是（n^3-n^2）/2
把n从2取到n,然后把每项相加、
得[(2^3+3^3+…+n^3)-(2^2+3^2+…+n^2)]/2
而2^3+3^3+…+n^3=(1+2+…+n)^2-1；2^2+3^2+…+n^2=n(n+1)(2n+1)/6-1
代入得={(1+2+…+n)^2-1-[n(n+1)(2n+1)/6-1]}/2=(3n^4+2n^3-3n^2-2n)/24
即(n－2)次幂的系数为(3n^4+2n^3-3n^2-2n)/24