2
$\begingroup$

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.

$\endgroup$
7
  • 1
    $\begingroup$ Let $a_n$ be the sequence, try simplifying $a_{n+1}-a_n$ $\endgroup$ – TravorLZH Jan 17 at 15:28
  • $\begingroup$ The method you used for finding it for product of 3 terms could be extended, couldn't it? $\endgroup$ – Light Yagami Jan 17 at 15:32
  • $\begingroup$ @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! $\endgroup$ – Cooperation Jan 17 at 15:37
  • $\begingroup$ Your first expression is undefined, as you don't say where is stops. $\endgroup$ – Yves Daoust Jan 17 at 15:37
  • $\begingroup$ @YvesDaoust He has written sum to $n$ terms... $\endgroup$ – Light Yagami Jan 17 at 15:38
3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $n=1$ $\endgroup$ – Henry Jan 17 at 15:42
  • $\begingroup$ Yes, that will be helpful for the OP. $\endgroup$ – S. Dolan Jan 17 at 15:44
1
$\begingroup$

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 $z^m$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $S_{n,m+1}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.