## Algebra 2 (1st Edition)

$50$
$\displaystyle \sum_{n=1}^{5}(n^{2}-1)=(1^{2}-1)+(2^{2}-1)+(3^{2}-1)+(4^{2}-1)+(5^{2}-1)$ $=(1-1)+(4-1)+(9-1)+(16-1)+(25-1)$ $=0+3+8+15+24$ $=50$