An interesting problem on progression and series 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.
 A: 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.
A: 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 $z^m$.
A: 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.
A: 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 $S_{n,m+1}$.
