## College Algebra (6th Edition)

$\displaystyle \sum_{k=1}^{15}ar^{k-1}$ (sample answer)
Remembering that $r^{0}=1$, we can write the first term as $ar^{0}$, (for k=1, the exponent is k-1) Observing the terms, counting them as k=2 (second), k=3 (third), etc, $ar^{1}$, (for k=2, the exponent is k-1) $ar^{2}$, (for k=3, the exponent is k-1) ... The pattern suggests that the general, k-th term is${}_{\text{ }} ar^{k-1}$ So, we have terms from 1 to 15, (the exponent in the last term is 14 = 15-1), and we write the sum as $\displaystyle \sum_{k=1}^{15}ar^{k-1}$ (sample answer)