Answer
The expression, $3\sum\limits_{k=1}^{5}{{{k}^{2}}}$, is easier to evaluate than the expression, $\sum\limits_{k=1}^{5}{3{{k}^{2}}}$.
Work Step by Step
$\sum\limits_{k=1}^{5}{3{{k}^{2}}\text{ }}\text{and 3}\sum\limits_{k=1}^{5}{{{k}^{2}}\text{ }}$
For the first expression,
$\begin{align}
& \sum\limits_{k=1}^{5}{3{{k}^{2}}\text{ }}=3\cdot {{1}^{2}}+3\cdot {{2}^{2}}+3\cdot {{3}^{2}}+3\cdot {{4}^{2}}+3\cdot {{5}^{2}} \\
& =3\cdot 1+3\cdot 4+3\cdot 9+3\cdot 16+3\cdot 25 \\
& =3+12+27+48+75 \\
& =165
\end{align}$
For the second expression,
$\begin{align}
& 3\sum\limits_{k=1}^{5}{{{k}^{2}}}=3\left( {{1}^{2}}+{{2}^{2}}+{{3}^{2}}+{{4}^{2}}+{{5}^{2}} \right) \\
& =3\left( 1+4+9+16+25 \right) \\
& =3\cdot 55 \\
& =165
\end{align}$
The second is easier to evaluate as the 3 is factored out.
