Intermediate Algebra for College Students (7th Edition)

$\sum_{k=1}^{(n+1)}[a+(k-1)d]$
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: $\sum_{k=1}^{(n+1)}[a+(k-1)d]$