If $m$ and $n$ are integers and $m\le n$, then
$$\dsum_{r=m}^na_r =a_m+a_{m+1}+a_{m+2}+\cdots+a_{n-1}+a_{n}\text{.}$$
Find the value of $\dsum_{i=2}^6 \dfrac{i^2}{3}$.
$\begin{array}{cl}
& \dsum_{i=2}^6 \dfrac{i^2}{3} \\
= & \dfrac{2^2}{3}+\dfrac{3^2}{3}+\dfrac{4^2}{3}+\dfrac{5^2}{3}+\dfrac{6^2}{3} \\
= & 30
\end{array}$
