Answer
$\sum_{k=1}^{(n+1)}[a+(k-1)d]$
Work Step by Step
Since, we have $a+(a+d)+(a+2d)+.....(a+nd)$
This form a arithmetic progression sequence whose first term is $a$ with common difference of $d$.
for this , we have
$a_n=a+(k-1)d$; for k-th term
Last term becomes $a+(k-1)d=a+nd$ or, $k=n+1$
Hence, $a+ar+ar^2+.....ar^{14}$ can be summed up with the summation notation as:
$\sum_{k=1}^{(n+1)}[a+(k-1)d]$
