每日一题[3380]三相循环

设 $\left(x^{2}+x-1\right)^{100}=a_{0}+a_{1} x+\cdots+a_{199} x^{199}+a_{200} x^{200}$，则

$$M=2 a_{0}-a_{1}-a_{2}+2 a_{3}-a_{4}-a_{5}+\cdots+2 a_{198}-a_{199}-a_{200}$$ 的值为（       ）

A．$2^{199}$

B．$2^{200}$

C．$2^{201}$

D．$2^{202}$

答案    C．

解析    根据题意，设 $\omega=\left(\dfrac{2\pi}3:1\right)$，则有

$$\begin{split} M&=2\sum_{k=0}^{66}\left(a_{3k}-\dfrac 12a_{3k+1}-\dfrac 12a_{3k+2}\right)\\ &=2\sum_{k=1}^{66}\left(\omega^{3k}a_{3k}+\omega^{3k+1}a_{3k+1}+\omega^{3k+2}a_{3k+2}\right)\\ &=2\sum_{k=0}^{200}\left(\omega^ka_k\right)\\ &=2\cdot \left(x^2+x-1\right)^{100}\Bigg|_{x=\omega}\\ &=2\left(\omega^2+\omega-1\right)^{100}\\ &=2\cdot (-2)^{100}\\ &=2^{201}. \end{split}$$