# Evaluating a Summation with a binomial

Problem:

Evaluate for n=11\begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} Sorry for this odd question. I saw this formula here on MSE which is quite helpful for a question I need to solve (\sin^{4n}\frac{\pi}{4n} + \cos^{4n} \frac{\pi}{4n})$for given values of$n$. Unfortunately I don't know how to evaluate this formula manually. I was thus hoping to use Wolfram Alpha to evaluate this for different values of$n$but was unable to enter it correctly. I would be truly grateful if somebody would kindly show me how to input this formula into Wolfram Alpha or solve it for$n=11. Many thanks in advance. • In general computer based systems have limitations. This appears to be one, in the general sense. The best reduction is to use the database of cosine values functions.wolfram.com/ElementaryFunctions/Cos/03/02 and write out the components of the series. There are also two typos and the formula should read \begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} Commented Jul 15, 2015 at 14:40 ## 3 Answers With some trivial manipulation it straighforward to check that we just have to compute: $$\sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\sum_{r=0}^{2n}\binom{4n}{2r}\exp\left(8r\cdot\frac{2\pi i}{4n}\right).\tag{1}$$ Since: $$\sum_{r=0}^{2n}\binom{4n}{2r} z^{2r} = \frac{1}{2}\sum_{k=0}^{4n}\binom{4n}{k}\left(z^k+(-z)^k\right)=\frac{(1+z)^{4n}+(1-z)^{4n}}{2}\tag{2}$$ by taking\omega=\exp\frac{2\pi i}{n}$we have: $$\sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n}=\text{Re}\left(\frac{(1+\omega)^{4n}+(1-\omega)^{4n}}{2}\right)\tag{3}$$ then, since$1\pm \omega = 2\cos\left(\frac{\pi}{n}\right)\exp\left(\pm\frac{\pi i}{n}\right), $$\sum_{r=0}^{2n}\binom{4n}{2r}\cos\frac{4\pi r}{n} = \frac{1}{2}\left(2\cos\frac{\pi}{n}\right)^{4n}\left(\cos(4\pi)+\cos(-4\pi)\right)=\color{red}{\left(2\cos\frac{\pi}{n}\right)^{4n}}.\tag{4}$$ • Thanks a lot for your efforts Sir. Sir I'm really sorry but there was a typo in the formula I had written. According to Leucippus Sir the formula should have been \begin{align} \sin^{4n}\left(\frac{\pi}{4n}\right) + \cos^{4n}\left( \frac{\pi}{4n}\right) = \frac{1}{4^{2n-1}} \left[ \sum_{r=0}^{n-1} \binom{4n}{2r} \cos\left(1 - \frac{r}{n} \right) \pi \, + \frac{1}{2} \binom{4n}{2n} \right]. \end{align} Sir please would it be possible for you to evaluate this series with anyn>11$as per your choice? Commented Jul 15, 2015 at 16:21 • @MakeaDifference: I am just too lazy to modify my answer accounting for your typo. My method works also in that case, you just have to take a different root of unity$\omega$. Commented Jul 15, 2015 at 16:32$\displaystyle e^{ix}=\cos x+i\sin x\implies e^{-ix}=\cos(-x)+i\sin(-x)=\cos x-i\sin x\displaystyle e^{ix}+e^{-ix}=2\cos x,e^{ix}-e^{-ix}=2i\sin x\displaystyle \left(2i\sin x\right)^{4n}+\left(2\cos x\right)^{4n}\displaystyle =(e^{ix}+e^{-ix})^{4n}+(e^{ix}-e^{-ix})^{4n}\displaystyle =2\sum_{r=0}^{2n}\binom{4n}{2r}(e^{ix})^{4n-2r}(e^{-ix})^{2r}\displaystyle\implies16^n\left(\sin^{4n}x+\cos^{4n}x\right)\displaystyle =2\sum_{r=0}^{2n}\binom{4n}{2r}e^{i4x(n-r)}\displaystyle 2^{4n-1}\left(\sin^{4n}x+\cos^{4n}x\right)=\binom{4n}{2n}+\sum_{r=0}^{n-1}\left(\binom{4n}{2r}e^{i4x(n-r)}+\binom{4n}{4n-2r}e^{-i4x(n-r)}\right)$As$\binom NR=\binom N{N-R},\displaystyle\implies2^{4n-1}\left(\sin^{4n}x+\cos^{4n}x\right)\displaystyle =\binom{4n}{2n}+\sum_{r=0}^n\binom{4n}{2r}\{e^{i4x(n-r)}+e^{-i4x(n-r)}\}\displaystyle\implies2^{4n-1}\left[\sin^{4n}x+\cos^{4n}x\right]=\binom{4n}{2n}+\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{4(n-r)x\}$Set$x=\dfrac\pi{4n}$Can you take it home from here? • Nomoshkar Sir. Thanks a lot Sir. Sir in the last result$$\displaystyle\implies16^n\left[\sin^{4n}x+\cos^{4n}x\right]=2\left[\binom {4n}{2n} 2+\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{(n-r)4x\}\right]$$$\left[\sin^{4n}x+\cos^{4n}x\right]$refers to the Greatest Integer Function right Sir? Also Sir please would you show me how to evaluate$2\left[\binom{4n}{2n}2+\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{(n-r)4x\}\right]?$I haven't yet learnt how to compute such series and thus resort to Wolfram Alpha for the computation. Unfortunately Sir I cannot understand how to input this series into Wolfram Alpha. Commented Jul 15, 2015 at 17:44 • Sir please would you evaluate the formula for any$n\ge 11?$I would feel really really grateful if you would kindly do this for me. Commented Jul 15, 2015 at 17:45 • @MakeaDifference, Please find the updated version. This holds true for any positive integer$n$. So, I'm not sure about specialty of$n\ge11$Commented Jul 16, 2015 at 5:44 • Sorry Sir but I'm still unable to evaluate the sum $$\sum_{r=0}^{n-1}\binom{4n}{2r}2\cos\{4(n-r)x\}$$ for$x=\dfrac\pi{4n}$and$n=11.$I've never learnt how to evaluate these types of series. I would be really grateful Sir if you would please evaluate the above series for me for$n=11?$Lastly Sir by $$\binom{4n}{2n}$$ do you mean $$\dfrac{44!}{22!22!}?$$ for$n=11?$Commented Jul 16, 2015 at 10:32 if you meant$\cos(1-4n/2r)*\pi$, then using Maple I got it seems for me that you cannot simplify the answer. If you meant$\cos[(1-4n/2r)*\pi]\$, then here is the result

• This is not an answer. Everyone is able to use a computer algebra system, but the point is to provide a manageable closed form for the given sum: it is not difficult to compute it through standard human-manipulations. Commented Jul 15, 2015 at 14:59
• try it first then do it by hand : ) I agree that we should not abuse computer alot! Commented Jul 15, 2015 at 15:04