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已知(x²+1)·(x-9)^9=a0+a1·(x-1)+a2·(x-1)²+……+a11·(x-1

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已知(x²+1)·(x-9)^9=a0+a1·(x-1)+a2·(x-1)²+……+a11·(x-1)^11
求(a1+3·a3+5·a5+……+11·a11)²-(2·a2+4·a4+6·a6+……+10·a10)²
题目(图片)http://hiphotos.baidu.com/%D1%A9%C9%BD%BD%A3%BF%CDty/pic/item/2d1d9d20b899a901c502818b1d950a7b0208f50e.jpg
令t=x-1 x=t+1
f(t)=(t^2+2t+2)(t-8)^9=a0+a1t……+a11t^11
f'(t)=(2t+2)(t-8)^9+9(t-8)^8(t^2+2t+2)=a1+2a2t+3a3t^2……+10a10t^9+11a11t^10
t=1 a1+2a2+……+11a11=17*7^8 (1)
t=-1 a1+2a2-3a3……+11a11=9^9 (2)
(a1+3·a3+5·a5+……+11·a11)²-(2·a2+4·a4+6·a6+……+10·a10)²
=(a1+2a2+3a3……+11a11)(a1-2a2+3a3-4a4……+11a11)
=17*7^8*9^9