## Intermediate Algebra for College Students (7th Edition)

$\sum_{i=1}^{15}(ar^{i-1})$
Since, we have $a+ar+ar^2+.....ar^{14}$ This form a geometric progression sequence whose first term is $a$ with common ratio of $r$. For a geometric progression sequence, we have $s_n=ar^{(n-1)}$ Last term becomes $ar^{(n-1)}=ar^{14}$ or, $n=15$ Hence, $a+ar+ar^2+.....ar^{14}$ can be summed up with the summation notation as: $\sum_{i=1}^{15}(ar^{i-1})$