## Algebra 2 (1st Edition)

$\displaystyle \sum_{i=1}^{8}(i+4)$
From the given terms, we observe the pattern $5=4+1$ $6=4+2$ $7=4+3$ $...$ $12=4+8$ So, there are 8 terms in all, counting each as $a_{i}, i=1,2,...8,$ (lower limit is 1, upper is 8) Sum = $\displaystyle \sum_{i=1}^{8}(i+4)$