Answer
$1260$
Work Step by Step
$2+4+6+\cdots+70=2+(2+2)+(2+2\cdot 2)+\cdots+(2+34\cdot 2)$
$4-2=2$
$6-4=2$
There is a constant difference, $2$, between terms.
The terms of the sum
are the first $35$ terms of an arithmetic sequence, $a_{1}=2, d=4.$
Sum of the First $n$ Terms of an arithmetic sequence:
$S_{n}=\displaystyle \frac{n}{2}\left(a_{1}+a_{n}\right)$
$S_{35}=\displaystyle \frac{35}{2}\left(2+70\right)$
$S_{35}=35\cdot 36=1260$
