Question

若(x平方+1)*[(x-2)的九次方]=a0+a1*(x-1)+a2*(x-1)平方+……+a11*(x-1)十一次方

Answer 1

左右分别求导

得到：

(11X^2-4X+9)(X-2)^8=a1+2a2(X-1)+3a3(X-1)^2+……＋11a11(X-1)*2

赋值X＝2,得到：

0＝a1+2a2+3a3+……+11a11

所求的式子作因式分解＝(a1+2a2+3a3+……+11a11)(a1-2a2+……-10a10+11a11)=0