Answer
True
Work Step by Step
$\sum\limits_{k=1}^{n}{c{{a}_{k}}}=c\sum\limits_{k=1}^{n}{{{a}_{k}}}$
$\begin{align}
& \sum\limits_{k=1}^{n}{c{{a}_{k}}}=c{{a}_{1}}+c{{a}_{2}}+c{{a}_{3}}+c{{a}_{4}}+\ldots +c{{a}_{n}} \\
& =c\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+{{a}_{4}}+\ldots +{{a}_{n}} \right) \\
& =c\sum\limits_{k=1}^{n}{{{a}_{k}}}
\end{align}$
Obtain the left side of the expression,
$\sum\limits_{k=1}^{n}{c{{a}_{k}}}=c\sum\limits_{k=1}^{n}{{{a}_{k}}}$
Thus, the statement is true.