# How do you find the sum of the series n^2 from n=0 to n=4?

Nov 22, 2016

${\sum}_{n = 0}^{4} {n}^{2} = 30$

#### Explanation:

We need the standard formula ${\sum}_{r = 1}^{n} {r}^{2} = \frac{1}{6} n \left(n + 1\right) \left(2 n + 1\right)$

$\therefore {\sum}_{n = 0}^{4} {n}^{2} = \frac{1}{6} \left(4\right) \left(4 + 1\right) \left(8 + 1\right)$
$\therefore {\sum}_{n = 0}^{4} {n}^{2} = \frac{1}{6} \left(4\right) \left(5\right) \left(9\right)$
$\therefore {\sum}_{n = 0}^{4} {n}^{2} = 30$

Alternatively, as there are only a few terms we could just write them out and compute the sum;
${\sum}_{n = 0}^{4} {n}^{2} = {0}^{2} + {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2}$
$\therefore {\sum}_{n = 0}^{4} {n}^{2} = 1 + 4 + 9 + 16$
$\therefore {\sum}_{n = 0}^{4} {n}^{2} = 30$, as before