Answer
True.
(see proof below)
Work Step by Step
$\displaystyle \sum_{i=1}^{n}ka_{i}=ka_{1}+ka_{2}+ka_{3}+ka_{4}+\cdots+ka_{n}$
... we can factor k out,
$=k(a_{1}+a_{2}+a_{3}+\cdots+a_{n})$
... and recognize the sum in the parentheses can be written in summation notation,
The lower limit is 1, and the upper limit is n.
$=k\displaystyle \cdot\sum_{i=1}^{n}a_{i}$
The statement is true.
