Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated. 

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sum_{k = 1}^{n}k\pars{k + 1}\pars{k + 2}\pars{k + 3}
&= \sum_{k = 0}^{n}\pars{k + 3}^{\underline{4}} \\
&= \left.{\pars{k + 3}^{\underline{5}} \over 5}\right\vert_{\ 0}^{\ n + 1} \\
&= {{\pars{n + 4}^{\underline{5}} \over 5} - {3^{\underline{5}} \over 5}}
\\[5mm] &=
\bbox[#ffe,20px,border:1px dotted navy]{\pars{n + 4}\pars{n + 3}\pars{n + 2}\pars{n + 1}n \over 5} \\ &
\end{align}

Note that $\ds{3^{\underline{5}} = 3 \times 2 \times 1 \times 0 \times \pars{-1} = {\large 0}}$.

Reference: See $\ds{\mathbf{2.50}\ \pars{~\mbox{page}\ 50~}}$ in Concrete Mathematics by R. L. Graham, D. E. Knuth and O. Patashnik, $\ds{\mrm{6^{th}}}$ ed. Addison-Wesley $\ds{1990}$.
