How to I solve this summation?

Question

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

Answer 1

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}