回復 13# anyway13 的帖子
\( \displaystyle (1+x)^{200} = \sum_{k=0}^{200} a_{k} x^{k} \)
令\( \displaystyle \omega = \cos{\frac{2\pi}{3}} + i\sin{\frac{2\pi}{3}} \; , \; 1 + \omega + \omega^2 = 0 \; , \; \omega^{3} = 1 \)
\( \displaystyle (1+1)^{200} = a_{0} + a_{1} + a_{2} + a_{3} + \cdots + a_{200} \)
\( \displaystyle (1+\omega)^{200} = a_{0} + a_{1} \omega + a_{2} \omega^2 + a_{3} \omega^3 + \cdots + a_{200} \omega^{200} \)
\( \displaystyle (1+\omega^2)^{200} = a_{0} + a_{1} \omega^2 + a_{2} \omega^4 + a_{3} \omega^6 + \cdots + a_{200} \omega^{400} \)
三式相加：
\( \displaystyle 2^{200} + (-\omega^2)^{200} + (-\omega)^{200} = 3( a_{0} + a_{3} + \cdots + a_{198} ) \)
\( \displaystyle \sum_{k=0}^{66} a_{3k} = \frac{ 2^{200} + \omega + \omega^2 }{3} = \frac{ 2^{200} - 1 }{3} \quad \Rightarrow \quad \sum_{k=1}^{66} a_{3k} = \frac{ 2^{200} - 1 }{3} - 1 = \frac{ 2^{200} - 4 }{3} \)。
