formula for $\left(1\cdot2\cdot...\cdot k\right)+...+\left(n\left(n+1\right)...\left(n+k- 1\right)\right)$

Question

proof if $k$ is a constant then for every $n$ natural numbers we have:

$$\left(1\cdot2\cdot...\cdot k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k- 1\right)\right)=\frac{n\left(n+1\right)\left(n+2\right)...\left(n+k\right)}{k+1}.$$ clearly $$\left(1\cdot2\cdot...\cdot k\right)+\left(2\cdot3\cdot...\cdot\left(k+1\right)\right)+...+\left(n\left(n+1\right)...\left(n+k- 1\right)\right)=\sum_{m=1}^{n}\left(\prod_{j=0}^{k-1}\left(m+j\right)\right)$$ $$=\sum_{m=1}^{n}\left(\Gamma\left(m+j+1\right)\right)$$where$\left(0\le j\le k-1\right)$ and $\Gamma(x)$ is Gamma function, or equivalently $$=\sum_{m=1}^{n}\int_{0}^{∞}x^{\left(m+j\right)}e^{-x}dx$$ but I could not find the given formula.

Answer 1

Note that by the Hockey-stick identity we have that $$1\cdot2\cdots k+2\cdot3\cdots(k+1)+\dots+n\cdot(n+1)\cdots(n+k-1)$$ $$\begin{align} &=k!\binom{k}{k}+k!\binom{k+1}{k}+\dots+k!\binom{n+k-1}{k}\\ &=k!\sum_{i=k}^{n+k-1}\binom{i}{k}\\ &=k!\binom{n+k}{k+1}\\ &=k!\cdot\frac{n\cdot(n+1)\cdots(n+k)}{(k+1)!}\\ &=\frac{n\cdot(n+1)\cdots(n+k)}{k+1}\\ \end{align}$$

Answer 2

We can prove this by induction. The base case here is when $n=1.$ Clearly,

$n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)=1\cdot(2)\cdot(3)\cdot\dots\cdot(k)$

and

$\dfrac{(n)\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k)}{1+k}=(1)\cdot(2)\cdot(3)\cdot\dots\cdot(k)=n\cdot(n+1)\cdot(n+2)\cdot\dots\cdot(n+k-1)$

so the base case holds. Assume the statement is true for some $y\in\mathbb{N}.$ Then

$$ \begin{align}(1)\cdot(2)\cdot\dots\cdot(k)+(2)\cdot(3)\cdot\dots\cdot(k+1)+\dots+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)=[(1)\cdot(2)\cdot\dots\cdot(k)+(2)\cdot(3)\cdot\dots\cdot(k+1)\\+(y)\cdot(y+1)\cdot\dots\cdot(y+k-1)]+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)\\=\dfrac{(y)\cdot(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)}{1+k}+(y+1)\cdot(y+2)\cdot\dots\cdot(y+k)=\dfrac{(y+1)\cdot(y+2)\dots(y+k+1)}{k+1},\end{align} $$

so the equation holds by induction.