# 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?

• – Lucian Nov 14 '14 at 18:16

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 $n$.

\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: $$\color{blue}{\int_{k=0}^{n+1}k^{r+1}dk}=\color{green}{\frac{(n+1)^{r+2}}{r+2}}$$