Showing $\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$ I was solving a question and saw a pattern. Can someone prove it, please?
We know
$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}$$
$$\sum_{k=1}^{n}k(k+1) = \frac{n(n+1)(n+2)}{3}$$
$$\sum_{k=1}^{n}k(k+1)(k+2) = \frac{n(n+1)(n+2)(n+3)}{4}$$
So we see the pattern... Can anyone give a proof of why:
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$
 A: This answer gives a proof by induction. Here instead I will give a proof using Hockey-stick identity:
$$\sum_{i=r}^n \binom ir = \binom{n+1}{r+1}$$
Note that
$$k(k+1)\cdots(k+r) = \frac {(k+r)!}{(k-1)!} = (r+1)!\binom{k+r}{r+1}$$
Hence your sum
$$=(r+1)!\sum_{k = 1}^{n} \binom {k+r}{r+1} =(r+1)!\sum_{k = {r+1}}^{n+r} \binom {k}{r+1} = (r+1)!\binom{n+r+1}{r+2} \\= \frac {(r+1)!(n+r+1)!}{(n-1)!(r+2)!} = \frac {n(n+1)\cdots(n+r+1)}{r+2}$$
A: You have
\begin{align}
k(k+1)(k+2)\cdots(k+r) &= k(k+1)(k+2)\cdots(k+r)\frac{r+2}{r+2}\\
&= k(k+1)(k+2)\cdots(k+r)\frac{(k+r+1)-(k-1)}{r+2}\\
&= \frac{k(k+1)(k+2)\cdots(k+r)(k+r+1)}{r+2}-\frac{(k-1)k(k+1)(k+2)\cdots(k+r)}{r+2}\\
\end{align}
Make the sum, you deduce easily that
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$
A: Let us count the number of subsets of $\{ 0, 1, \dots, n + r\}$ of size $r+2$ in two different ways. One way, of course, is $$\binom{n+r+1}{r+2} = \frac{n(n+1)\cdots (n+r+1)}{(r+2)!}.\tag{1}$$
For the other way, let us first pick the largest element that will be in our set, and call it $j$. Thus, the remaining $r+1$ elements are chosen from $\{0, 1, \dots, j-1\}$. Summing over all values of $j$, we obtain $$\sum_{j=r+1}^{n+r} \binom{j}{r+1}.$$ Writing $k = j - r$, we find that this sum equals $$\sum_{k=1}^n \binom{k+r}{r+1} = \sum_{k=1}^n\frac{k(k+1)\cdots(k+r)}{(r+1)!}.\tag{2}$$
Equating $(1)$ and $(2)$ and cancelling $(r+1)!$ in the denominator, we obtain the result.
