<s>[INST] 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:
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$
 [/INST] 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
$$\sum_{k=1}^{n}k(k+1)(k+2)\cdots(k+r) = \frac{n(n+1)(n+2)\cdots(n+r+1)}{r+2}$$
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