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. 
 A: If I were you, I would try induction. First establish that it is true for $n=1$:
$$1*2*3*4=24=\frac{1(2)(3)(4)(5)}{5}$$
Then use induction by first assuming that, for some $k$,
$$1*2*3*4+...+k(k+1)(k+2)(k+3)=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$
Then add $(k+1)(k+2)(k+3)(k+4)$ to both sides:
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}+(k+1)(k+2)(k+3)(k+4)$$
If you want to try this on your own, stop reading here and try the induction yourself.
Factor out $(k+1)(k+2)(k+3)$ from the right side:
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)(\frac{k(k+4)}{5}+(k+4))$$
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=(k+1)(k+2)(k+3)(\frac{(k+5)(k+4)}{5})$$
$$1*2*3*4+...+(k+1)(k+2)(k+3)(k+4)=\frac{(k+1)(k+2)(k+3)(k+4)(k+5)}{5}$$
And there you have it.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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 $\ds{1990}$.
A: $$\sum_{k=1}^n k(k+1)(k+2)(k+3)$$
$$=\sum_{k=1}^n 4!\binom{k+3}{4}$$
$$=4!\sum_{k=1}^n \binom{k+3}{4}$$
Using the hockey-stick identity:
$$=4!\binom{k+4}{5}$$
$$=\frac{k(k+1)(k+2)(k+3)(k+4)}{5}$$
A: HINT:
From the Right Hand Side,  if $f(n)=\dfrac{n(n+1)(n+2)(n+3)(n+4)}5$
$$f(m+1)-f(m)=?$$
A: There is a mistake here:
$$1^4+2^4+\cdots+n^4=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}$$
It should be
$$1^4+2^4+\cdots+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
You approach has no problem, except this minor mistake.
A: On your last equation, putting $n=-2,-3,-4$ leads the equation equal to zero.  Then $n=1$ and the value of $\frac{1*2*3*4*5}5$ leads $\frac{n(n+1)(n+2)(n+3)(n+4)}5$.
