How to find the sum of $i(i+1)\cdots(i+k)$ for fixed $k$ between $i = 1$ and $n$? I learned that $$\sum \limits_{i=1}^n i(i+1) = \frac{n(n+1)(n+2)}{3}$$ or in general $$\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \frac{n(n+1)\dots (n+k+1)}{k+2}$$ From a mathematical standpoint why is this true? I'm not asking for inductive proof. I am asking if you only given the left hand side, how would you go about writing a closed form expression for the sum?
 A: $$
S=\sum \limits_{i = 1}^n i(i+1)(i+2) \dots (i + k) = \sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!}
$$
$$
\frac{S}{(k+1)!}=\sum_{i=1}^{n}\frac{(i+k)!}{(i-1)!(k+1)!}=\sum_{i=0}^{n-1}\binom{i+k+1}{i}
$$
$$
\frac{S}{(k+1)!}=\binom{k+1}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\
=\binom{k+2}{0} + \binom{k+2}{1} + \dots + \binom{n+k}{n-1} \\
=\binom{k+3}{1} + \binom{k+3}{2} + \dots + \binom{n+k}{n-1} \\
=\binom{n+k+1}{n-1}
$$
the above uses $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$
$$
S=\binom{n+k+1}{n-1}*(k+1)!=\frac{(n+k+1)!(k+1)!}{(n-1)!(k+2)!}= \frac{n(n+1)\dots (n+k+1)}{k+2}
$$
A: There is a combinatorial argument (used several times in this site) which explains these identities: 
$$
\sum \limits_{i = 1}^n i(i+1)(i+2) \cdots (i + k) = \frac{n(n+1)\dots (n+k+1)}{k+2}
$$
Rather it explains THESE equivalent identities:
$$
\sum \limits_{i = 1}^n \binom{i+k}{k+1}=\frac{1}{(k+1)!}\sum \limits_{i = 1}^n i(i+1)(i+2) \cdots (i + k) = \frac{n(n+1)\dots (n+k+1)}{(k+2)!}=\binom{n+k+1}{k+2}
$$
So, the right-hand side of the above is the number $N$ of ways (or combinations) we can pick $k+2$ elements from the set $\{1,2,\ldots,n+k+1\}$. This number $N$ can be split as
$$
N=N_{k+2}+N_{k+3}+\cdots+N_{n+k+1},
$$
where $N_{k+i}$ is the number of those previous combinations where the largest number in the combination is $k+i$, and hence $N_{k+i}$ is equal to the number of ways we pick $k+1$ elements from the set $\{1,2,\ldots,k+i-1\}$, and thus
$$
N_{k+i}=\binom{k+i-1}{k+1}.
$$ 
Thus
$$
\binom{k+1}{k+1}+\binom{k+2}{k+1}+\cdots+\binom{n+k}{k+1}=\binom{n+k+1}{k+2}.
$$
A: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\isdiv}{\,\left.\right\vert\,}
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$\ds{\sum_{i = 1}^{n}i\pars{i + 1}\pars{i + 2}\ldots\pars{i + k}
     ={n\pars{n + 1}\ldots\pars{n + k + 1} \over k + 2}:\ {\large ?}}$

\begin{align}
&\sum_{\ell = 1}^{n}\ell\pars{\ell + 1}\pars{\ell + 2}\ldots\pars{\ell + k}=
\sum_{i = 1}^{n}{\pars{\ell + k}! \over \pars{\ell - 1}!}
=\pars{k + 1}!\sum_{\ell = 1}^{n}{\ell + k \choose k + 1}
\\[3mm]&=\pars{k + 1}!\sum_{\ell = 1}^{n}
\int_{\verts{z} = 1}{\pars{1 + z}^{\ell + k} \over z^{k + 2}}\,{\dd z \over 2\pi\ic}
=\pars{k + 1}!\int_{\verts{z} = 1}{\pars{1 + z}^{k} \over z^{k + 2}}
\sum_{\ell = 1}^{n}\pars{1 + z}^{\ell}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{k + 1}!\int_{\verts{z} = 1}{\pars{1 + z}^{k} \over z^{k + 2}}\,
{\pars{1 + z}\bracks{\pars{1 + z}^{n} - 1} \over \pars{1 + z} - 1}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\pars{k + 1}!\
\overbrace{\int_{\verts{z} = 1}
{\pars{1 + z}^{k + 1 + n} \over z^{k + 3}}\,{\dd z \over 2\pi\ic}}
^{\ds{=\ {n + k + 1 \choose k + 2}}}\
-\
\pars{k + 1}!\
\overbrace{\int_{\verts{z} = 1}{\pars{1 + z}^{k + 1} \over z^{k + 3}}\,{\dd z \over 2\pi\ic}}^{\ds{=\ 0}}
\\[3mm]&=\pars{k + 1}!\,{\pars{n + k + 1}! \over \pars{k + 2}!\pars{n - 1}!}
=\color{#f00}{%
\pars{k + 1}!}\,{\pars{n + k + 1}\ldots\pars{n + 1}n\,\color{#f00}{\pars{n - 1}!}
\over \pars{k + 2}\color{#f00}{\pars{k + 1}!}\,\color{#f00}{\pars{n - 1}!}}
\end{align}

$$
\color{#00f}{\large\sum_{i = 1}^{n}i\pars{i + 1}\pars{i + 2}\ldots\pars{i + k}
={n\pars{n + 1}\ldots\pars{n + k + 1} \over k + 2}}
$$
