I am trying to prove that $$\sum_{k=0}^{\lfloor\frac{n-1}{2}\rfloor}\binom{n}{2k+1}(-1)^k = \sqrt{2^n}\sin{}n\pi/4$$

This follows from the subtraction of $\sum_{k=0}^{...}\binom{n}{4k+1}$ and $\sum_{k=1}^{...}\binom{n}{4k-1}$, which according to wikipedia can be evaluated using the so called multisection of sums for binomial coefficients: https://en.wikipedia.org/wiki/Binomial_coefficient#Multisections_of_sums. Is there a simple proof for the identity provided in the link? Or is there a different proof for my identity of which I'm not aware of? Thanks

  • $\begingroup$ Sorry if I'm wrong, but maybe this seems like a good place to use some properties of the gamma function, more specifically the euler's reflection formula? Because there's some sort of $\sin$ and factorials involved. $\endgroup$ – Mathsisfun May 30 '20 at 11:06
  • $\begingroup$ Note that we can get rid of $\sin$ if we consider $n=4m, n=4m+1, \cdots$ $\endgroup$ – Peanut May 30 '20 at 11:23

Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$ we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$ Now take $a_k=\binom{n}{k+1}i^k$, where $i=\sqrt{-1}$, to obtain \begin{align} \sum_{k\ge 0} \binom{n}{2k+1}(-1)^k &= \sum_{k\ge 0} \binom{n}{2k+1}i^{2k} \\ &= \sum_{k\ge 0} \binom{n}{k+1}i^k \frac{1+(-1)^k}{2} \\ &= \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1}i^k + \frac{1}{2}\sum_{k\ge 0} \binom{n}{k+1} (-i)^k \\ &= \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1}i^{k+1} - \frac{1}{2i}\sum_{k\ge 0} \binom{n}{k+1} (-i)^{k+1} \\ &= \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k}i^k - \frac{1}{2i}\sum_{k\ge 1} \binom{n}{k} (-i)^k \\ &= \frac{1}{2i}\left((1+i)^n-1\right) - \frac{1}{2i}\left((1-i)^n-1\right) \\ &= \frac{(1+i)^n - (1-i)^n}{2i} \\ &= \frac{\left(\sqrt{2}(\cos(\pi/4)+i\sin(\pi/4)\right)^n - \left(\sqrt{2}(\cos(-\pi/4)+i\sin(-\pi/4)\right)^n}{2i} \\ &= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(-n\pi/4)+i\sin(-n\pi/4))}{2i} \\ &= \frac{\sqrt{2}^n(\cos(n\pi/4)+i\sin(n\pi/4)) - \sqrt{2}^n(\cos(n\pi/4)-i\sin(n\pi/4))}{2i} \\ &= \sqrt{2}^n \sin(n\pi/4) \end{align}


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