Answer
$\displaystyle \sum_{k=1}^{n}(a+d^{k})$
(sample answer)
Work Step by Step
(first term) : a+$d^{1}$
(2nd term): a+$d^{2}$
(3rd term): a+$d^{3}$
...
If we begin counting (indexing) terms with $k=1$,
the pattern suggests that the k-th term is built by
adding $d^{k}$ to a$:$
$a_{k}= a+d^{k}$
The last term is such that we added $d^{n}$ to a,
meaning that k=n.
We began with $k=1$, and ended with $k=n$:
$\displaystyle \sum_{k=1}^{n}(a+d^{k})$
(sample answer)
