How to I solve this summation? 
*

*I am having trouble solving this summation: 
$\displaystyle{\quad\sum_{i = 1}^{n}\,\,\sum_{j = 4}^{i}
\left(\,\, j + 2i\,\right)}$.

*I've only gotten this far:
$\displaystyle{\quad\sum_{i = 1}^{n}\sum_{j = 4}^{i}2i +
{i\,\left(\, i + 4\,\right) \over 2}\quad}$ and would welcome some help.

 A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\color{#f00}{\sum_{k = 1}^{n}\,\,\sum_{j = 4}^{k}\pars{j + 2k}} & =
\sum_{k = 1}^{n}\,\,\sum_{j = 1}^{k - 3}\pars{j + 3 + 2k} =
\sum_{k = 1}^{n}
\bracks{{\pars{k - 3}\pars{k - 2} \over 2} + \pars{k - 3}\pars{2k + 3}}
\\[4mm] & =
{5 \over 2}\sum_{k = 1}^{n}k^{2} - {11 \over 2}\sum_{k = 1}^{n}k -
6\sum_{k = 1}^{n}1
\\[4mm] & =
{5 \over 2}\,{n\pars{n + 1}\pars{2n + 1} \over 6} -
{11 \over 2}\,{n\pars{n + 1} \over 2} - 6n =
\color{#f00}{{5 \over 6}\,n^{3} - {3 \over 2}\,n^{2} - {25 \over 3}\,n}
\end{align}
