Answer
$\sum_{i=1}^{n}2i$
Work Step by Step
We contract the sum into sigma notation by noting that it follows the pattern of a sum of even integers (e.g. $2i$), starting with $2$ (e.g. $i=1$), and ending with $2n$ (e.g. $i=n$).
$2+4+6+8+\displaystyle \cdots+2n=\sum_{i=1}^{n}2i$
