# How do you evaluate the sum represented by sum_(n=1)^(10)n^2 ?

${\sum}_{k = 1}^{n} {k}^{2} = \frac{n \left(n + 1\right) \left(2 n + 1\right)}{6}$,
${\sum}_{n = 1}^{10} {n}^{2} = \frac{\left(10\right) \left(11\right) \left(21\right)}{6} = 385$