Answer
True.
$\displaystyle \sum_{i=1}^{4}(i^2+2i)=\displaystyle \sum_{i=1}^{4}i^2+\displaystyle \sum_{i=1}^{4}2i$
Work Step by Step
$\displaystyle \sum_{i=1}^{4}(i^2+2i)=1^2+2(1)+2^2+2(2)+3^2+2(3)+4^2+2(4)=(1^2+2^2+3^2+4^2)+(2(1)+2(2)+2(3)+2(4))=\displaystyle \sum_{i=1}^{4}i^2+\displaystyle \sum_{i=1}^{4}2i$