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In my text book I found a problem which asked to sum of the following series $$1\times 2\times 3+2\times 3 \times 4+ 3\times 4\times 5+\cdots + n(n+1)(n+2)$$ which I found to be $\frac{n(n+1)(n+2)(n+3)}{4}$ which is indeed true.

Now as a general thought it came to my mind what if the problem asks to find the closed form of this$-$ $$\sum_{k=1}^{n}\underbrace{k\times(k+1)\times(k+2)\times(k+3)\times(k+4)\cdots\times(k+m-1)}_{m \text { terms}}$$ where $m$ is any integer for instance in the aforementioned problem it was $4$. Now, following the pattern our intuition suggest that the answer should be $$\boxed{\frac{n(n+1)(n+2)\cdots(n+m)}{m+1}}$$ Now, in order to check one can easily say that for $m=1$ it's obviously true. It's also true for $m=2$. So now my natural question is that Is the closed sum true for all $m \in\mathbb{N}$ ? I have tried with induction but at the end I mess it up and the idea to proof in that way seems to be went into vain and I also have no further idea to proceed, any help? Thanks for your attention.

Cooperation
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    Let $a_n$ be the sequence, try simplifying $a_{n+1}-a_n$ – TravorLZH Jan 17 '21 at 15:28
  • The method you used for finding it for product of 3 terms could be extended, couldn't it? – V.G Jan 17 '21 at 15:32
  • @Light I tried using general pattern of the terms and then using the result I just summed it up but that became exceeding difficult for higher terms...I mean I think so! – Cooperation Jan 17 '21 at 15:37
  • Your first expression is undefined, as you don't say where is stops. –  Jan 17 '21 at 15:37
  • @YvesDaoust He has written sum to $n$ terms... – V.G Jan 17 '21 at 15:38
  • @LightYagami: right, my bad. –  Jan 17 '21 at 15:38
  • Note that the general term of the sum is $a_k=\dfrac{(k+m-1)!}{(k-1)!}=m!\dbinom{k+m-1}m$ and you can use Pascal's rule and telescopy to note that $\sum_1^n a_k=m!\dbinom{n+m}{m+1}=\dfrac{n(n+1)\cdots (n+m)}{m+1}$ – Prasun Biswas Jan 17 '21 at 15:40

4 Answers4

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The result that $$\frac{n(n+1)(n+2)\cdots(n+m)}{m+1}-\frac{(n-1)n(n+1)\cdots(n+m-1)}{m+1}$$ $$=n(n+1)\cdots(n+m-1)$$ can be seen easily by taking out (mentally) the common factor $n(n+1)\cdots(n+m-1)$ of all terms.

You can then prove the result either by induction or by the method of differences.

  • If you use this for an induction proof, you might also say it is obvious that $\frac{n(n+1)(n+2)\cdots(n+m)}{m+1}=n(n+1)(n+2)\cdots(n+m-1)$ when $n=1$ – Henry Jan 17 '21 at 15:42
  • Yes, that will be helpful for the OP. –  Jan 17 '21 at 15:44
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The product $$ \prod\limits_{k = 0}^{m - 1} {\left( {z + k} \right)} = z^{\,\overline {\,m\,} } = {{\Gamma \left( {z + m} \right)} \over {\Gamma \left( z \right)}} = \left( {z + m - 1} \right)^{\,\underline {\,m\,} } $$

is known as the Rising factorial and is in general defined for $z,m \in \mathbb C$.
Here we will consider for simplicity the case $m \in \mathbb Z$, and $z^{\,\underline {\,m\,} } ,\quad z^{\,\overline {\,m\,} } $ will represent respectively the Falling and Rising Factorial.

One of its properties is that the Finite Difference (unitary step) is $$ \Delta _{\,z} \;z^{\,\overline {\,m\,} } = \left( {z + 1} \right)^{\,\overline {\,m\,} } - z^{\,\overline {\,m\,} } = m\left( {z + 1} \right)^{\,\overline {\,\,m - 1\,\,} } $$ or $$ \Delta _{\,z} \;\left( {z - 1} \right)^{\,\overline {\,m + 1\,} } = z^{\,\overline {\,m + 1\,} } - \left( {z - 1} \right)^{\,\overline {\,m + 1\,} } = \left( {m + 1} \right)z^{\,\overline {\,\,m\,\,} } $$

Then the sum has a clean and straight formulation $$ \eqalign{ & \sum\limits_{k = 1}^n {\left( {z + k} \right)^{\,\overline {\,m\,} } } = {1 \over {m + 1}}\sum\limits_{k = 1}^n {\Delta _{\,z} \;\left( {z + k - 1} \right)^{\,\overline {\,m + 1\,} } } = \cr & = {1 \over {m + 1}}\sum\limits_{k = 0}^{n - 1} {\Delta _{\,z} \;\left( {z + k} \right)^{\,\overline {\,m + 1\,} } } = {1 \over {m + 1}}\sum\limits_{k = 0}^{n - 1} {\left( {\left( {z + 1 + k} \right)^{\,\overline {\,m + 1\,} } - \left( {z + k} \right)^{\,\overline {\,m + 1\,} } } \right)} = \cr & = {1 \over {m + 1}}\left( {\left( {z + n} \right)^{\,\overline {\,m + 1\,} } - z^{\,\overline {\,m + 1\,} } } \right) \cr} $$

That is, the Falling/Rising factorials have difference / sum results that are the discrete analog of the derivative /integral results for $z^m$.

G Cab
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Suppose we want to evaluate the sum which you have already found out.

We evaluate the sum $$\displaystyle \sum_{k=1}^n k(k+1)(k+2)$$ Now if we multiply up and down by $4$, and write $4$ as $(k+3)-(k-1)$ (why? To make it a telescopic sequence!)

We get the sum as $$\dfrac{1}4 \left[\sum_{k=1}^n k(k+1)(k+2)(k+3)-\sum_{k=1}^n (k-1)(k)(k+1)(k+2)\right]$$

Now it is clear that it telescopes...so this method can be extended for higher terms as well.

V.G
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Let your sum be denoted $S_{n,m}$. If we increment $m$, the last factor in every term is $k+m=(k-1)+(m+1)$. This gives you two new sums, namely $S_{n-1,m+1}$ (by shifting the index, the first terms being $0$) and $(m+1)S_{n,m}$. Hence the recurrence

$$S_{n,m+1}=S_{n-1,m+1}+(m+1)S_{n,m}$$

from which

$$S_{n,m}=\frac{S_{n,m+1}-S_{n-1,m+1}}{m+1}.$$

The numerator is the last term of $S_{n,m+1}$.