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Let $x \in \mathbb{R}$ and let $n \in \mathbb{N}$. Evaluate: $$\sum_{k=0}^n \binom{n}{k} \sin\bigg(x+\frac{k\pi}{2}\bigg)$$ How to start it?

My aatempt: Let us assume that $n$ is odd. Then we have $$\sum_{k=0}^n \binom{n}{k} \sin\bigg(x+\frac{k\pi}{2}\bigg) = C_0 \sin x +C_1 \cos x - C_2 \sin x - C_3 \cos x + \ldots + C_{n-1}\sin x + C_n \cos x $$ How to proceed?

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  • $\begingroup$ Wonder if it would help to take two derivatives and compare to the original. You may get a DE you can recognize. $\endgroup$ – MPW Jun 9 '18 at 11:22
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With the help of complex number.

The result is equivalent to the imaginary part of $e^{ix}(1+e^{i\frac{\pi}{2}})^n$.

$e^{ix}(1+e^{i\frac{\pi}{2}})^n=e^{ix}(1+i)^n=e^{ix}2^{{n}/{2}}e^{in\pi/4}$

So the result is $2^{{n}/{2}}\sin(x+n\pi/4)$.

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  • $\begingroup$ Brilliant. Very nice and easy way $\endgroup$ – Mittal G Jun 9 '18 at 11:28

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