Answer
$2160$
Work Step by Step
$S=15+(15+2)+(15+4) +\cdots+(15+(39(2)))$
The terms of the sum form an arithmetic sequence with
$d=2,a_{1}=15,$ and $n=40$ .
The sum $S_{n}$ of the first $n$ terms of $\left\{a_{n}\right\}$:
$S_{n}=\displaystyle \frac{n}{2}\left[2a_{1}+(n-1)d\right]$
$S_{40}=\displaystyle \frac{40}{2}\left[2(15)+2(39)\right]=20(108)=2160$
