一个组合求和题

转化与化归

化简: $\sum_{k=0}^n(-1)^k \mathrm{C}_n^k \cos (k+1) \alpha$

kuing

总感觉做复杂了……
\(\newcommand\relph{\mathrel{\phantom=}}\)

记 $p=\cos x+i\sin x$，则
\[\sum_{k=0}^n(-1)^kC_n^k\cos\bigl((k+1)x\bigr)
=\sum_{k=0}^n(-1)^kC_n^k\frac{p^{k+1}+p^{-k-1}}2
=\frac12p(1-p)^n+\frac12p^{-1}(1-p^{-1})^n,\]
因为
\begin{align*}
1-p&=1-\cos x-i\sin x
=2\sin \frac x2\left( \sin \frac x2-i\cos \frac x2 \right)
=2\sin \frac x2\left( \cos \frac{x-\pi}2+i\sin \frac{x-\pi}2 \right), \\
1-p^{-1}&=1-\cos x+i\sin x
=2\sin \frac x2\left( \sin \frac x2+i\cos \frac x2 \right)
=2\sin \frac x2\left( \cos \frac{\pi-x}2+i\sin \frac{\pi-x}2 \right),
\end{align*}
于是
\begin{align*}
\frac12p(1-p)^n+\frac12p^{-1}(1-p^{-1})^n
&=\frac12(\cos x+i\sin x)\cdot 2^n\sin^n\frac x2\left( \cos \frac{n(x-\pi)}2+i\sin \frac{n(x-\pi)}2 \right) \\
&\relph {}+\frac12(\cos x-i\sin x)\cdot 2^n\sin^n\frac x2\left( \cos \frac{n(\pi-x)}2+i\sin \frac{n(\pi-x)}2 \right) \\
&=2^n\sin^n\frac x2\left( \cos \frac{n(x-\pi)}2\cos x -\sin \frac{n(x-\pi)}2\sin x \right) \\
&=2^n\sin^n\frac x2\cos \left( \frac{n(x-\pi)}2+x \right),
\end{align*}
所以最终化简结果为
\[\sum_{k=0}^n(-1)^kC_n^k\cos\bigl((k+1)x\bigr)
=2^n\sin^n\frac x2\cos \frac{(n+2)x-n\pi}2.\]

战巡

回复 2# kuing


令
\[S(x)=\sum_{k=0}^n(-1)^kC^k_n\cos((k+1)x)+i\sum_{k=0}^n(-1)^kC^k_n\sin((k+1)x)\]
\[=\sum_{k=0}^n(-1)^kC^k_ne^{ix}=e^{ix}(1-e^{ix})^n\]
\[=(\cos(x)+i\sin(x))(2\sin(\frac{x}{2})(\cos(\frac{x-\pi}{2})+i\sin(\frac{x-\pi}{2}))^n\]
\[=2^n\sin^n(\frac{x}{2})(\cos(x)+i\sin(x))(\cos(n\frac{x-\pi}{2})+i\sin(n\frac{x-\pi}{2}))\]
而
\[\sum_{k=0}^n(-1)^kC^k_n\cos((k+1)x)=Re(S(x))\]
\[=2^n\sin^n(\frac{x}{2})[\cos(x)\cos(n\frac{x-\pi}{2})-\sin(x)\sin(n\frac{x-\pi}{2})]=...\]

转化与化归

回复 2# kuing
牛！解得妙！

转化与化归

回复 3# 战巡
牛！牛！牛！

kuing

回复 3# 战巡

nice! 这个确实更简洁些，而且还顺便直接得出了 sin 的

PS、第二行有笔误，和式那里是 $e^{i(k+1)x}$