# How do you find the sum of the series k^2 from k=1 to 35?

${\sum}_{k = 1}^{35} {k}^{2} = 14910$
We need the standard formula ${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$
$\therefore {\sum}_{k = 1}^{35} {k}^{2} = \frac{1}{6} \left(35\right) \left(35 + 1\right) \left(70 + 1\right)$
$\therefore {\sum}_{k = 1}^{35} {k}^{2} = \frac{1}{6} \left(35\right) \left(36\right) \left(71\right)$
$\therefore {\sum}_{k = 1}^{35} {k}^{2} = 14910$