Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$

Considering the following formulae:

(i) $1+2+3+..+n = n(n+1)/2$

(ii) $1\cdot2+2\cdot3+3\cdot4+…+n(n+1) = n(n+1)(n+2)/3$

(iii) $1\cdot2\cdot3+2\cdot3\cdot4+…+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4$

Find and prove a ‘closed formula’ for the sum

$1\cdot2\cdot3\cdot…\cdot k + 2\cdot3\cdot4\cdot…\cdot(k+1) + … + n(n+1)(n+2)\cdot…\cdot (k+n-1)$

generalizing the formulae above.

I have attempted to ‘put’ the first 3 formulae together but I am getting nowhere and wondered where to even start to finding a closed formula.

Solutions Collecting From Web of "Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$"

The pattern looks pretty clear: you have

\begin{align*} &\sum_{i=1}^ni=\frac12n(n+1)\\ &\sum_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)\\ &\sum_{i=1}^ni(i+1)(i+2)=\frac14n(n+1)(n+2)(n+3)\;, \end{align*}\tag{1}

where the righthand sides are closed formulas for the lefthand sides. Now you want

$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\;;$$

what’s the obvious extension of the pattern of $(1)$? Once you write it down, the proof will be by induction on $n$.

Added: The general result, of which the three in $(1)$ are special cases, is $$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\frac1{k+1}n(n+1)(n+2)\dots(n+k)\;.\tag{2}$$ For $n=1$ this is $$k!=\frac1{k+1}(k+1)!\;,$$ which is certainly true. Now suppose that $(2)$ holds. Then

\begin{align*}\sum_{i=1}^{n+1}i(i+1)&(i+2)\dots(i+k-1)\\ &\overset{(1)}=(n+1)(n+2)\dots(n+k)+\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)\\ &\overset{(2)}=(n+1)(n+2)\dots(n+k)+\frac1{k+1}n(n+1)(n+2)\dots(n+k)\\ &\overset{(3)}=\left(1+\frac{n}{k+1}\right)(n+1)(n+2)\dots(n+k)\\ &=\frac{n+k+1}{k+1}(n+1)(n+2)\dots(n+k)\\ &=\frac1{k+1}(n+1)(n+2)\dots(n+k)(n+k+1)\;, \end{align*}

exactly what we wanted, giving us the induction step. Here $(1)$ is just separating the last term of the summation from the first $n$, $(2)$ is applying the induction hypothesis, $(3)$ is pulling out the common factor of $(n+1)(n+2)\dots(n+k)$, and the rest is just algebra.

If you divide both sides by $k!$ you will get binomial coefficients and you are in fact trying to prove
$$\binom kk + \binom{k+1}k + \dots + \binom{k+n-1}k = \binom{k+n}{k+1}.$$
This is precisely the identity from this question.

The same argument for $k=3$ was used here.

Or you can look at your problem the other way round: If you prove this result about finite sums
$$\sum_{j=1}^n j(j+1)\dots(j+k-1)= \frac{n(n+1)\dots{n+k-1}}{k+1},$$
you also get a proof of the identity about binomial coefficients.

For a fixed non-negative $k$, let $$f(i)=\frac{1}{k+1}i(i+1)\ldots(i+k).$$
Then $$f(i)-f(i-1)=i(i+1)\ldots(i+k-1).$$
By telescoping,

$$\sum_{i=1}^ni(i+1)(i+2)\dots(i+k-1)=\sum_{i=1}^n\left(f(i)-f(i-1)\right)=f(n)-f(0)=f(n)$$

and we are done.

I asked exactly this question a couple of days ago, here:

Telescoping series of form $\sum (n+1)\cdot…\cdot(n+k)$

My favourite solution path so far is
$$n(n+1)\cdot…\cdot(n+k)/(k+1)$$