Answer
$\sum_{i=1}^{13}(ar^{i-1})$
Work Step by Step
Since, we have $a+ar+ar^2+.....ar^{12}$
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 $ar^{(n-1)}=ar^{12}$ or, $n=13$
Hence, $a+ar+ar^2+.....ar^{12}$ can be summed up with the summation notation as:
$\sum_{i=1}^{13}(ar^{i-1})$
