Answer
$$\frac{46168}{3}$$
Work Step by Step
Given $$\sum_{j=2}^{30}\left(6 j+\frac{4 j^{2}}{3}\right)$$
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_{j=2}^{30}\left(6 j+\frac{4 j^{2}}{3}\right)&= 6 \sum_{j=1}^{30} j-6+\frac{4}{3} \sum_{j=1}^{30} j^{2}-\frac{4}{3} ,\ \ \text{Use (1), ( 2)}\\
&= 6\left(\frac{30(30+1)}{2} \right)-6+\frac{4}{3}\left(\frac{30(30+1)(2 (30)+1)}{6}\right)-\frac{4}{3}\\
&=\frac{46168}{3}
\end{align*}
