Answer
$\displaystyle \sum_{k=1}^{15}ar^{k-1}$
(sample answer)
Work Step by Step
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)
