Question

高中数学 二项式定理 详细解释一下

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）https://gss0.baidu.com/7LsWdDW5_xN3otqbppnN2DJv/%D1%A9%C9%BD%BD%A3%BF%CDty/pic/item/2d1d9d20b899a901c502818b1d950a7b0208f50e.jpg

Answer 1

设f(x)=(x^2+1)(x-9)=a0+a1(x-1)+a2(x-1)^2+…+a11(x-11)^11。
f'(x)=(x^2+1)'(x-9)+(x^2+1)(x-9)'=3x^2-18x+1
=a1+2a2(x-1)+3a3(x-1)^2+4a4(x-1)^3+5a5(x-1)^4+…+10a10(x-1)^9+11a11(x-1)^10
f'(-1)=3*(-1)^2-18*(-1)+1=22
=a1-2a2+3a3-4a4+…-10a10+11a11
=(a1+3a3+5a5+…+11a11)-(2a2+4a4+6a6+…+10a10)
所以，(a1+3a3+5a5+…+11a11)-(2a2+4a4+6a6+…+10a10)=22。

Answer 2

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

Answer 3

f（x）
f‘（x）=2x（x-9）^9+9(x^2+1)(x-9)^8=a1+2a2(x-1)+...+11a11(x-1)^10
f'(2)=a1+2a2+...+11a11=17*7^8
f'(0)=a1-2a2+...-10a10+11a11=9^9
所以（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