Answer
$\displaystyle \sum_{k=1}^{n}ar^{k-1}$
or $\displaystyle \sum_{k=0}^{n-1}ar^{k}$
Work Step by Step
The first term (when k=$1$) is $a=ar^{0}=ar^{1-1}$,
The second term (when k=$2$) is $ar=ar^{2-1}$,
The third term (when k=$3$) is $ar^{2}=ar^{3-1}$,
The kth is $ar^{k-1}$.
The last term is for $k=n$, and equals $ar^{n-1}$
So
$a+ar+ar^{2}+\displaystyle \dots+ar^{n-1}=\sum_{k=1}^{n}ar^{k-1}$
or if we count with k from 0 to n-1,
$=\displaystyle \sum_{k=0}^{n-1}ar^{k}$
