Sum of Squares
$1+2^2+3^2+\cdots+n^2 = \frac{1}{3}n(n+1)\left(n+\frac{1}{2}\right)=\frac{n(n+1)(2n+1)}{6}$
Basel's Problem
$$1+\frac{1}{4}+\frac{1}{9}+\cdots = \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi ^2}{6}$$
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