# Evaluating $\sum_{n = 1}^{71} \sin^{2560} \left(\frac{n\pi}{71}\right)$

I want to evaluate the following summation:
$$\sum_{n = 1}^{71} \sin^{2560}\left(\frac{n\pi}{71}\right)$$ The way I approached this problem was to convert $\sin\left(\frac{n\pi}{71}\right)$ into complex form using: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ Substitute $\sin(x)$ with complex form and solve the series to finally deduce to: $$\frac{71}{2^{2560}}\sum_{n=0}^{36}\left(-1\right)^n \begin{pmatrix} 2560\\71n+2\\\end{pmatrix}$$ This is where I finally stuck but I tried to use calculator with lower power of $\sin$ (i.e. $\sin^{50}\left(\frac{n\pi}{23}\right)$ or $\sin^{20}\left(\frac{n\pi}{71}\right)$) to calculate the alternative but similar version of the original summation to see how would the answer looks like, so far, the answer seems to be $$\frac{71}{2^{2560}}\begin{pmatrix} 2560\\1280\\\end{pmatrix}$$ though this have yet to be confirmed or proven.
The summation I have got make it difficult to confirm the answer since it chooses very specific terms to compute i.e. every 71 terms and it is also alternating.
Which method should I approach this problem and is my approach appropriated? If so then how can I solve for combinatorial sum of this kind?

• I think your sum $\sum_{n=0}^{36}\dots$ is correct, but not the final formula. Jun 29, 2017 at 20:14
• The first is $1.07593$ and the last is $1.11953$. Jun 30, 2017 at 0:43
• Is there supposed to be a neat closed-form answer for this summation? Jun 30, 2017 at 0:51

This sum is

$$S = \sum_{k=0}^{p-1} \sin^{4n}(k \pi/p)$$

where $n=640$ and $p=71.$ In what follows we will take $p$ odd.

We also have

$$\sum_{k=p}^{2p-1} \sin^{4n}(k \pi/p) = \sum_{k=0}^{p-1} \sin^{4n}((k+p) \pi/p) = \sum_{k=0}^{p-1} \sin^{4n}(k \pi/p)$$

so we are in fact dealing with

$$\frac{1}{2} \sum_{k=0}^{2p-1} \sin^{4n}(k \pi/p) = \frac{1}{2} \sum_{k=0}^{2p-1} \sin^{4n}(2k \pi/2/p).$$

Introducing

$$f(z) = \frac{1}{2} \frac{(z-1/z)^{4n}}{(2i)^{4n}} \frac{2p z^{2p-1}}{z^{2p}-1} = \frac{1}{2} \frac{1}{2^{4n} (-1)^{2n}} \frac{(z^2-1)^{4n}}{z^{4n}} \frac{2p z^{2p-1}}{z^{2p}-1} \\ = \frac{p}{2^{4n}} \frac{(z^2-1)^{4n}}{z^{4n+1-2p}} \frac{1}{z^{2p}-1}$$

we get

$$S = \sum_{k=0}^{2p-1} \mathrm{Res}_{z=2\pi ik/2/p} f(z).$$

This implies

$$S = - \mathrm{Res}_{z=\infty} f(z) - \mathrm{Res}_{z=0} f(z).$$

We heve for the contribution from the first residue $$\mathrm{Res}_{z=0} \frac{1}{z^2} f(1/z).$$

This yields

$$\mathrm{Res}_{z=0} \frac{1}{z^2} \frac{p}{2^{4n}} (1/z^2-1)^{4n} z^{4n+1-2p} \frac{1}{1/z^{2p}-1} \\ = \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{p}{2^{4n}} \frac{(1-z^2)^{4n}}{z^{8n}} z^{4n+1} \frac{1}{1-z^{2p}} \\ = \mathrm{Res}_{z=0} \frac{1}{z^{4n+1}} \frac{p}{2^{4n}} (1-z^2)^{4n} \frac{1}{1-z^{2p}}.$$

This is

$$[z^{4n}] \frac{p}{2^{4n}} (1-z^2)^{4n} \frac{1}{1-z^{2p}} = [z^{2n}] \frac{p}{2^{4n}} (1-z)^{4n} \frac{1}{1-z^{p}}.$$

We thus obtain

$$\frac{p}{2^{4n}} \sum_{q=0}^{\lfloor 2n/p\rfloor} (-1)^{2n-pq} {4n\choose 2n-pq}.$$

We get for the contribution from the second residue

$$- [z^{4n-2p}] \frac{p}{2^{4n}} (z^2-1)^{4n} \frac{1}{z^{2p}-1} = [z^{2n-p}] \frac{p}{2^{4n}} (z-1)^{4n} \frac{1}{1-z^{p}}$$

which is

$$\frac{p}{2^{4n}} \sum_{q=0}^{\lfloor (2n-p)/p\rfloor} (-1)^{4n-(2n-p-pq)} {4n\choose 2n-p-pq}.$$

We thus have combining the two pieces (with $p$ odd we have $(-1)^{2n-pq} = (-1)^q$ and $(-1)^{2n+p+pq} = - (-1)^{pq} = - (-1)^q$)

$$\frac{p}{2^{4n}} \sum_{q=0}^{\lfloor 2n/p\rfloor} (-1)^{q} {4n\choose 2n-pq} - \frac{p}{2^{4n}} \sum_{q=0}^{\lfloor 2n/p\rfloor - 1} (-1)^{q} {4n\choose 2n-p-pq} \\ = \frac{p}{2^{4n}} \sum_{q=0}^{\lfloor 2n/p\rfloor} (-1)^{q} {4n\choose 2n-pq} - \frac{p}{2^{4n}} \sum_{q=1}^{\lfloor 2n/p\rfloor} (-1)^{q-1} {4n\choose 2n-pq}.$$

This yields the end result

$$\bbox[5px,border:2px solid #00A000]{ \frac{p}{2^{4n}} {4n\choose 2n} + 2 \times \frac{p}{2^{4n}} \sum_{q=1}^{\lfloor 2n/p\rfloor} (-1)^{q} {4n\choose 2n-pq}.}$$

In the specific case being asked this works out to

$$\bbox[5px,border:2px solid #00A000]{ \frac{71}{2^{2560}} {2560\choose 1280} + 2 \times \frac{71}{2^{2560}} \sum_{q=1}^{18} (-1)^{q} {2560\choose 1280-71q}.}$$