Question

若多项式X+X^11=a0+a1(x-1)+...+a10(x-1)^10+a11(x-1)^11,则a10等于多少

Answer 1

f(x)=(x^2+1)(x-2)^9=a0+a1(x-1)+a2(x-1)^2+...+a11(x-1)^11
f(x)=a0+a1(x-1)+a2(x-1)^2+...+a11(x-1)^11对x求导，
就是
f'(x)=a1+2a2(x-1)+3a3(x-1)^2+...+11a11(x-1)^10
f(x)=(x^2+1)(x-2)^9对x求导
同时f'(x)=2x(x-2)^9+9(x^2+1)(x-2)^8
可知，f'(2)=a1+2a2+...+11a11=0
f'(0)=a1+3a3+……+11a11)-(2a2+4a4+……+10a10)
所以(a1+3a3+……+11a11)^2-(2a2+4a4+……+10a10)^2
=[(a1+3a3+……+11a11)+(2a2+4a4+……+10a10)]*[(a1+3a3+……+11a11)-(2a2+4a4+……+10a10)]
=f'(2)*f'(0)=0.

Answer 2

比较不等式两边x^11的系数
1=a11;
比较不等式两边x^10的系数
0=a10+（-1）^1
a11(这里用到了二项式定理）;
a10=11a11=11
C