Answer
$\text {(a)}$
\begin{align}
\sum_{k=0}^{14} (k+4)(k+1)
\end{align}
$\text {(b)}$
\begin{align}
\sum_{k=5}^{19} (k-1)(k-4)
\end{align}
Work Step by Step
$\text {The given summation is}$
\begin{align}
\sum_{k=4}^{18} k(k-3)
\end{align}
k = 0 should be the lower limit:}$
\begin{align}
k' = k -4 \Rrightarrow k = k' + 4 \\
when \ k' = 0 \Rrightarrow k = 4 \\
when \ k' = 14 \Rrightarrow k = 18 \\
k \Rrightarrow k + 4 \ and \ k - 3 \Rrightarrow k+1
\end{align}
$\text {Therefore, the summation will be}$
\begin{align}
\sum_{k=0}^{14} (k+4)(k+1)
\end{align}
$\text {(b) k = 5 should be the lower limit:}$
\begin{align}
k' = k +1 \Rrightarrow k = k' -1 \\
when \ k' = 5 \Rrightarrow k = 4 \\
when \ k' = 19 \Rrightarrow k = 18 \\
k \Rrightarrow k -1\ and \ k - 3 \Rrightarrow k-4
\end{align}
$\text {Therefore, the summation will be}
\begin{align}
\sum_{k=5}^{19} (k-1)(k-4)
\end{align}
