Intuition behind sum of multiplication arithmetic sequence Maybe this is a stupid question but please guide and enlighten me patiently. I have just known something fact that quite shocking me. Let start from this simple fact
$$\sum_{k=1}^n k=\frac{n(n+1)}{2}\tag{1}$$
The summation above is a sum of arithmetic progression with common difference of 1 and I have already known it. Then, it turns out (I realized these when playing with Wolfram|Alpha)
$$\begin{align}\sum_{k=1}^n k(k+1)&=\frac{n(n+1)(n+2)}{3}\tag{2}\\\sum_{k=1}^n k(k+1)(k+2)&=\frac{n(n+1)(n+2)(n+3)}{4}\tag{3}\\\sum_{k=1}^n k(k+1)(k+2)(k+3)&=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}\tag{4}\\\end{align}$$
and it seems (I haven't proved it yet)
$$\sum_{k=1}^n k(k+1)(k+2)\cdots(k+r)=\frac{n(n+1)(n+2)(n+3)(n+4)\cdots(n+r+1)}{r+2}\tag{5}$$
We have an obvious pattern here. I know the intuition of $(1)$, but I am wondering what are the intuitutions for the other sums: $(2),\,(3),\,(4),\,(5)$?
I can derive $(2)$ using well-known formulas for arithmetic series and square pyramidal number, but how do the other formulas, $(3),\,(4),\,(5)$, derive? Does it use Faulhaber's formula?
 A: The derivation is by straightforward induction on $n$. Suppose
$\sum_{k=1}^n k(k+1)...(k+r)=\frac{n(n+1)(n+2)...(n+r+1)}{r+2}$
Then
$$\begin{align}
\sum_{k=1}^{n+1} k(k+1)...(k+r) & =\frac{n(n+1)(n+2)...(n+r+1)}{r+2} + (n+1)(n+2)...(n+r+1) \\
& = (n+1)(n+2)...(n+r+1)\left(\frac{n}{r+2} + 1\right) \\
& = (n+1)(n+2)...(n+r+1)\left(\frac{n + r + 2}{r+2}\right) \\
& =\frac{(n+1)(n+2)...(n+r+1)(n+r+2)}{r+2}
\end{align}$$
Now you just have to check the base case $n=1$ to establish the formula for all $n$.
A: $$\begin{align}
\sum_{k=1}^{n+1}k(k+1)\cdots(k+r)&=\sum_{k=1}^{n+1}k^\overline{r+1}\\
&=\frac1{r+2}\sum_{k=1}^{n+1}\left[k^\overline{r+2}-(k-1)^\overline{r+2}\right]\\
&=\frac1{r+2}\left[\sum_{k=1}^{n+1}k^\overline{r+2}-\sum_{k=0}^{n}k^\overline{r+2}\right]\\
&=\frac{(n+1)^\overline{r+2}}{r+2}\\
&=\frac{(n+1)(n+2)\cdots(n+r+2)}{r+2}\qquad \blacksquare
\end{align}$$
Alternative method, using binomial coefficients:
Note that
$$\sum_{j=b}^{m}\binom {a+j}{a+b}=\binom{a+m+1}{a+b+1}$$
Hence
$$\begin{align}
\sum_{k=1}^{n+1}k(k+1)\cdots(k+r&)=(r+1)!\sum_{k=1}^{n+1}\binom{k+r}{r+1}\\
&=(r+1)!\binom{n+r+2}{r+2}\\
&=(r+1)!\frac{(n+r+2)^\underline{r+2}}{(r+2)(r+1)!}\\
&=\frac{(n+1)^\overline{r+2}}{r+2}\qquad \blacksquare
\end{align}$$

NB:
It is interesting to note the striking resemblance of the result above to a power integral.
Result above (taking summation from $0$ instead of $1$):
$$\color{blue}{\sum_{k=0}^{n+1}k^\overline{r+1}}=\color{green}{\frac{(n+1)^\overline{r+2}}{r+2}}$$
Power integral:
$$\color{blue}{\int_{k=0}^{n+1}k^{r+1}dk}=\color{green}{\frac{(n+1)^{r+2}}{r+2}}$$
