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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3What does $C_i$ represent? – Gregoire Rocheteau Dec 01 '18 at 04:46
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They are the binomial coefficients. $C_i=(^n_i)$ – Shubham Johri Dec 01 '18 at 05:07
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I will point out that some sums of this type are mentioned in Wikipedia article on binomial coefficients - in the section Multisections of sums - however without a proof. (I will also add link to the current revision.) – Martin Sleziak Dec 01 '18 at 08:41
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And you can also find several posts on this site which are about similar sums (although perhaps not the same) - looking at the solutions there might help you. For example: Find the value $\binom {n}{0} + \binom{n}{4} + \binom{n}{8} + \cdots $, where $n$ is a positive integer., Finding $\binom{n}{0} + \binom{n}{3} + \binom{n}{6} + \ldots $ or How do I count the subsets of a set whose number of elements is divisible by 3? 4? – Martin Sleziak Dec 01 '18 at 08:43
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https://math.stackexchange.com/questions/3016886/calculate-sum-limits-n-0-infty-fracx3n3n/3016891#3016891 – lab bhattacharjee Dec 05 '18 at 12:30
1 Answers
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$(1+x)^n=C_0+C_1x+C_2x^2+...+C_nx^n\\\implies(1+1)^n=2^n=C_0+C_1+...+C_n\\\implies(1-1)^n=0=C_0-C_1+...+(-1)^nC_n\\\implies(1+i)^n=2^{n/2}e^{n\pi/4}=(C_0-C_2+C_4...)+i(C_1-C_3+C_5...)\\\ \ \ \ \ \ \ \ \ \ \ \ 2^{n/2}\sin n\pi/4=C_1-C_3+C_5...$
$\frac{2^n-0}2+2^{n/2}\sin n\pi/4=2(C_1+C_5...)$
Shubham Johri
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