# How do you find the sum of the series k^2-1 from k=1 to 5?

Nov 10, 2016

${\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = 50$

#### Explanation:

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

$\therefore {\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = {\sum}_{k = 1}^{5} \left({k}^{2}\right) - {\sum}_{k = 1}^{5} \left(1\right)$
$\therefore {\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = \frac{1}{6} \left(5\right) \left(5 + 1\right) \left(10 + 1\right) - 5$
$\therefore {\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = \frac{1}{6} \left(5\right) \left(6\right) \left(11\right) - 5$
$\therefore {\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = 55 - 5$
$\therefore {\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = 50$

Alternatively, as there are only a few terms we could just write them out and compute the sum;
${\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = \left(1 - 1\right) + \left(4 - 1\right) + \left(9 - 1\right) + \left(16 - 1\right) + \left(25 - 1\right)$
$\therefore {\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = 0 + 3 + 8 + 15 + 24$
$\therefore {\sum}_{k = 1}^{5} \left({k}^{2} - 1\right) = 50$, as before