From Hirakata-city, Osaka-prefecture, Japan

I wrote a formula using mathjax. For the format, this site will be helpful \sum_{ k = 1 }^{ n } k^2 = \overbrace{ 1^2 + 2^2 + \cdots + n^2 }^{ n } = \frac{ 1 }{ 6 } n ( n + 1 ) ( 2n + 1 )  $$\begin{array}{r}\sum _{k=1}^n{k}^2=\stackrel{n}{\overbrace{1^2+2^2+\cdots +n^2}}=\frac{1}{6}n(n+1)(2n+1)\end{array}$$