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Prove that $$C_1+C_5+ C_9+... =\frac{1}{2}\bigg(2^{n-1}+2^{n/2}\sin\frac{n\pi}{4}\bigg)$$

Here $C_i$ denotes the binomial coefficient $\binom ni$.

I tried to solve this problem by using de Moivre's theorem but could not proceed further.

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closed as off-topic by Saad, Cesareo, Brahadeesh, I am Back, Did Dec 1 '18 at 17:11

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$(1+x)^n=C_0+C_1x+C_2x^2+...+C_nx^n\\\implies(1+1)^n=2^n=C_0+C_1+...+C_n\\\implies(1-1)^n=0=C_0-C_1+...+(-1)^nC_n\\\implies(1+i)^n=2^{n/2}e^{n\pi/4}=(C_0-C_2+C_4...)+i(C_1-C_3+C_5...)\\\ \ \ \ \ \ \ \ \ \ \ \ 2^{n/2}\sin n\pi/4=C_1-C_3+C_5...$

$\frac{2^n-0}2+2^{n/2}\sin n\pi/4=2(C_1+C_5...)$

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