Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$ Simplify the following $(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))$.
How do i do this by using these identities.
$C^r_r + C^{r+1}_r + C^{r+2}_r + ....+C^{n}_r = C^{n+1}_{r+1}$
Or
$C^r_0 + C^{r+1}_1 + C^{r+2}_2 + ....+C^{r+k}_k = C^{r+k+1}_{k}$
 A: Hint:
$$(1\times2)+(2\times3)+(3\times4)+\dots(n\times(n+1)) =\sum_{k=1}^nk(k+1) \\
\begin{align}
& =2\sum_{k=1}^n\frac{k(k+1)}2 \\
& =2\sum_{k=1}^nC_2^{k+1} \\
\end{align}$$
A: Note that $2\binom{k}{2}=(k-1)k$. Hence
$$(1\cdot2)+(2\cdot3)+(3\cdot4)+\dots+(n\cdot(n+1))=\sum_{k=2}^{n+1}2\binom{k}{2}=2\binom{n+2}{3}.$$
A: $\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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 \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}}}
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\begin{align}
&\color{#f00}{\pars{1 \cdot 2} + \pars{2 \cdot 3} + \pars{3 \cdot 4} + \cdots + \bracks{n\pars{n + 1}}} =
\sum_{k = 1}^{n}k\pars{k+1} = \sum_{k = 1}^{n}\bracks{k\pars{k - 1} + 2k}
\\[5mm] = &\
\sum_{k = 1}^{n}\pars{k^{\underline{2}} + 2k^{\underline{1}}}
\qquad\qquad
\pars{~a^{\underline{n}} \equiv a\pars{a - 1}\ldots\pars{a - n + 1}\
\mbox{is the}\ falling\!-\!factorial~}
\\[5mm] = &\
\left.{1 \over 3}\,k^{\underline{3}} + k^{\underline{2}}
\,\right\vert_{\ 1}^{\  n + 1} =
{1 \over 3}\,\pars{n + 1}^{\underline{3}} + \pars{n + 1}^{\underline{2}} -
{1 \over 3}\,1^{\underline{3}} - 1^{\underline{2}}
\\[5mm] = &\
{1 \over 3}\pars{n + 1}n\pars{n - 1} + \pars{n + 1}n =
\color{#f00}{{1 \over 3}\pars{n + 2}\pars{n + 1}n} =
{1 \over 3}\,n^{3} + n^{2} + {2 \over 3}\,n
\end{align}
A: This doesn't answer using your combinatorics argument, but I figured I would provide an arguably easier solution in case you just want an answer. 
What you have is basically the sum 
$$\sum_{k=0}^n k(k+1)$$. 
You can break this up into two sums, 
$$\sum_{k=0}^n k^2+\sum_{k=0}^n k$$
Now, we have well known solutions for these sums,  which are,  respectively, $\frac{n(n+1)(2n+1)}{6}$ and $\frac{n(n+1)}{2}$. Adding those together will yield your answer. 
