Question

The sum $\sum _{m=1}^{4n+1}{\left[\sum _{k=1^m}^{m+1}(\sin \left(\frac{2\pi k}{m}\right)-i\cos \left(\frac{2\pi k}{m}\right))\right]}^m$ is

• A. independent of $n$
• B. purely imaginary
• C. purely real
• D. a root of $x^{4n+1}+1=0$

Answer 1

Answer: (A), (B)

The correct options are
A independent of $n$
B purely imaginary

Let $S=\sum _{m=1}^{4n+1}{\left[\sum _{k=1^m}^{m+1}\left\{\sin \left(\frac{2\pi k}{m}\right)-i\cos \left(\frac{2\pi k}{m}\right)\right\}\right]}^m$

Let $t_k=\sin \left(\frac{2\pi k}{m}\right)-i\cos \left(\frac{2\pi k}{m}\right)\Rightarrow t_k=-i\left\{i\sin \left(\frac{2\pi k}{m}\right)+\cos \left(\frac{2\pi k}{m}\right)\right\}$

Assume $\alpha =i\sin \left(\frac{2\pi }{m}\right)+\cos \left(\frac{2\pi }{m}\right)$ .....(1)

$\Rightarrow t_k=-i{\alpha }^k$

Now $S_1=\sum _{k=1}^{m+1}{t}_k=-i\sum _{k=1}^{m+1}{\alpha }^k=-i(\alpha +{\alpha }^2+{\alpha }^3+....+{\alpha }^{m+1})$

$\Rightarrow S_1=-i\left(\frac{{\alpha }^m-1}{\alpha -1}\right)$

From (1), we have

${\alpha }^m=\cos 2\pi +i\sin 2\pi =1$

So, $\Rightarrow S_1=i\left(\frac{\alpha -1}{\alpha -1}\right)=1$

So, $\Rightarrow S=\sum _{n=1}^{4n+1}{S}_1=\sum _{n=1}^{4n+1}{i}^m=i\frac{(i^{4n+1}-1)}{i-1}=i$

So, options A and B are correct.