A series problem.

Viewed 82

I personally encountered a series problem that I have no idea.

$$
\sum^n_{r=0} \binom{n}{r} \sin[(2r+1)\alpha]
$$

where $\alpha \in (-\frac{1}{2}\pi,\frac{1}{2}\pi)$

My personal guess is that this could probably relate to complex numbers but I don't have a clear idea.

He is

Are you chinese?

1 Answers

$$\begin{aligned}S &= \text{Im}\left[ \sum_{r=0}^{n} \binom{n}{r} e^{i(2r+1)\alpha} \right] \\
&= \text{Im}\left[ e^{i\alpha} \sum_{r=0}^{n} \binom{n}{r} (e^{i2\alpha})^r \right]\\
&= \text{Im}\left[ e^{i\alpha} (1 + e^{i2\alpha})^n \right]\\
&=2^n \cos^n \alpha \sin(n+1)\alpha\\
\end{aligned}$$

反馈一个bug:你的回答可以被我编辑

因为这是初创的社区。所以给了用户很大的权限。

Related

互联网ICP备案:沪ICP备2025152146号

© 2025 任务优先(上海)网络科技有限公司 保留所有权利。