What's an intuitive explanation of the following mathematical fact: [math]\displaystyle \sum_{n=1}^{\infty}{\frac{1}{n^2}} = \frac{\pi^2}{6}[/math]? - Quora
![Prove that sum(r=0)^n^n Cr(-1)^r[i+i^(2r)+i^(3r)+i^(4r)]=2^n+2^(n/2+1)cos(npi//4),w h e r ei=sqrt(-1)dot Prove that sum(r=0)^n^n Cr(-1)^r[i+i^(2r)+i^(3r)+i^(4r)]=2^n+2^(n/2+1)cos(npi//4),w h e r ei=sqrt(-1)dot](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/32132_web.png)
Prove that sum(r=0)^n^n Cr(-1)^r[i+i^(2r)+i^(3r)+i^(4r)]=2^n+2^(n/2+1)cos(npi//4),w h e r ei=sqrt(-1)dot
![sequences and series - Using roots of unity to prove that $\cos\frac{\pi }{2n}\cos\frac{2\pi}{2n}\cdots\cos\frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$ - Mathematics Stack Exchange sequences and series - Using roots of unity to prove that $\cos\frac{\pi }{2n}\cos\frac{2\pi}{2n}\cdots\cos\frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$ - Mathematics Stack Exchange](https://i.stack.imgur.com/plfM7.jpg)
sequences and series - Using roots of unity to prove that $\cos\frac{\pi }{2n}\cos\frac{2\pi}{2n}\cdots\cos\frac{(n-1)\pi}{2n}=\frac{\sqrt{n}}{2^{n-1}}$ - Mathematics Stack Exchange
![sequences and series - Are there any visual proofs for $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$? - Mathematics Stack Exchange sequences and series - Are there any visual proofs for $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$? - Mathematics Stack Exchange](https://i.stack.imgur.com/tC7aY.png)