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:
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$

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
$$=(r+1)!\sum_{k = 1}^{n} \binom {k+r}{r+1} =(r+1)!\sum_{k = {r+1}}^{n+r} \binom {k}{r+1} = (r+1)!\binom{n+r+1}{r+2} \\= \frac {(r+1)!(n+r+1)!}{(n-1)!(r+2)!} = \frac {n(n+1)\cdots(n+r+1)}{r+2}$$
