Answer
$$\frac{20215}{2}$$
Work Step by Step
Given $$\sum_{m=1}^{20}\left(5+\frac{3 m}{2}\right)^{2}$$
Since
\begin{align}
\sum_{j=1}^{N} j&=1+2+\cdots+N=\frac{N(N+1)}{2} \tag{1}\\
\sum_{j=1}^{N} j^{2}&=1^{2}+2^{2}+\cdots+N^{2}=\frac{N(N+1)(2 N+1)}{6} \tag{2}\\
\sum_{j=1}^{N} j^{3}&=1^{3}+2^{3}+\cdots+N^{3}=\frac{N^{2}(N+1)^{2}}{4} \tag{3}
\end{align}
Then
\begin{align*}
\sum_{m=1}^{20}\left(5+\frac{3 m}{2}\right)^{2}&= \sum_{m=1}^{20}\left(25+15 m+\frac{9 m^{2}}{4}\right)\\
&=\sum_{m=1}^{20} 25+\sum_{m=1}^{20} 15 m+\sum_{m=1}^{20} \frac{9 m^{2}}{4}\\
&=25(20)+15 \sum_{m=1}^{20} m+\frac{9}{4} \sum_{m=1}^{20} m^{2}\ \ \text{Use (1), (2)}\\
&=25(20)+15\frac{20(20+1)}{2} +\frac{9}{4} \frac{20(20+1)(2(20)+1)}{6} \\
&=\frac{20215}{2}
\end{align*}
