Telescoping series of form $\sum (an+1) ..(an+k)$ If we are given a series say
$$\sum_{n=1}^N (n+1)(n+2) ... (n+k)$$ we can find a telescoping series by noting that
$(n+1)(n+2)..(n+k)(n+k+1) - n(n+1)..(n+k) = (n+k+1-n) (n+1)...(n+k)$
and hence able to write
$$\sum_{n=1}^N (n+1)(n+2) ... (n+k) = \frac{1}{k+1}\sum_{n=1}^N (n+1)...(n+k+1) - n(n+1)...(n+k)$$
However, for series like $$\sum_{n=1}^N (2n+1)(2n+2) ... (2n+k)$$ or for a general multiple of $n$, such as $$\sum_{n=1}^N (an+1)(an+2) ... (an+k)$$ the same trick to add a term before and after would not work.
Are there any ways to express series like this as a telescoping sum, or failing that, any useful (relatively) closed form identities? If so, a pointer to a reference would be much appreciated! 
 A: The original identity comes from finding an antidifference for the function $n^{\underline{k}}$.  Unfortunately, no such antidifference exists for $(2n)^{\underline{k}}$, due to lack of a "chain rule" for antidifferences.  However here's an approach that you might like:
$$a_k=\sum_{n=1}^N(2n+1)(2n+2)\cdots(2n+k)$$
$$b_k=\sum_{n=1}^N (2n)(2n+1)\cdots(2n+k-1)$$
Multiplying the summands of $a_k$ by $(2n+k+1-2n)=k+1$, we get $a_k=\frac{1}{k+1}(a_{k+1}-b_{k+1})$.  Multiplying the summands of $b_k$ by $(2n+k-2n+1)=k+1$, we get $b_k=\frac{1}{k+1}(b_{k+1}-\sum_{n=0}^{N-1}(2n+1)(2n+2)\cdots(2n+k+1))=\frac{1}{k+1}(b_{k+1}-a_{k+1}+s_k)$, where $s_k=(2N+1)(2N+2)\cdots(2N+k+1)-(k+1)!$.
Adding, we get $$a_k+b_k=\frac{s_k}{k+1}$$
Replacing $k$ by $k+1$ and rearranging, we get $$-b_{k+1}=a_{k+1}-\frac{s_{k+1}}{k+2}$$
We plug into our above formula for $a_k$ to get $a_k=\frac{1}{k+1}(2a_{k+1}-\frac{s_{k+1}}{k+2})$.  Now we may solve for $a_{k+1}$ to get
$$a_{k+1}=\frac{k+1}{2}a_k+\frac{s_{k+1}}{2k+4}$$
which admittedly is no closed form, but it is a first-order recurrence.
A: Let
$P(k,a,j) =\prod_{i=1}^k (ai+j)
=a^k \prod_{i=1}^k (i+j/a)
$.
Instead of comparing
the next term with the current term,
I will compare the term
$a$ further on
(which is the next term
when $a=1$).
$\begin{align}
P(k, a, j+a) - P(k, a, j)
&=a^k \prod_{i=1}^k (i+(j+a)/a)
-a^k \prod_{i=1}^k (i+j/a)\\
&=a^k\big( \prod_{i=1}^k (i+(j+a)/a)
-\prod_{i=1}^k (i+j/a)\big)\\
&=a^k\big( \prod_{i=1}^k (i+1+j/a)
-\prod_{i=1}^k (i+j/a)\big)\\
&=a^k\big( \prod_{i=2}^{k+1} (i+j/a)
-\prod_{i=1}^k (i+j/a)\big)\\
&=a^k \big(\prod_{i=2}^{k} (i+j/a)\big)
\big ((k+1+j/a)
- (1+j/a)\big)\\
&=a^k k\prod_{i=2}^{k} (i+j/a)\\
&=a k\prod_{i=2}^{k} (ai+j)\\
&=a k\prod_{i=1}^{k-1} (a(i+1)+j)\\
&=a k\prod_{i=1}^{k-1} (ai+a+j)\\
&=a kP(k-1, a, j+a)\\
\end{align}
$
or
$P(k+1, a, j+a) - P(k+1, a, j)
=a(k+1) P(k, a, j+a)
$
or
$\dfrac{P(k+1, a, j+a) - P(k+1, a, j)}{a(k+1)} 
= P(k, a, j+a)
$
or
$\dfrac{P(k+1, a, j) - P(k+1, a, j-a)}{a(k+1)} 
= P(k, a, j)
$
For $a=1$
this becomes
$P(k+1, 1, j+1) - P(k+1, 1, j)
=(k+1)P(k-1, 1, j+1)
$
or
$P(k+1, 1, j) - P(k+1, 1, j-1)
=(k+1)P(k-1, 1, j)
$.
So
(note the limits of summation)
$\begin{align}
\sum_{j=1}^{Na} P(k, a, j)
&=\frac1{a(k+1)}\sum_{j=1}^{Na} \big(P(k+1, a, j) - P(k+1, a, j-a)\big)\\
&=\frac1{a(k+1)}\big(\sum_{j=1}^{Na} P(k+1, a, j) -\sum_{j=1}^{Na} P(k+1, a, j-a)\big)\\
&=\frac1{a(k+1)}\big(\sum_{j=Na-a+1}^{Na} P(k+1, a, j) -\sum_{j=1-a}^{0} P(k+1, a, j)\big)\\
\end{align}
$
We have $a$ terms which involve $N$
and $a$ terms which do not
(and so are constant).
If the upper limit is not a multiple of $a$,
then there are from $1$ to
$a-1$ terms
of the form
$P(k, a, Na+i)$.
This is a reasonable generalization
(IMHO) of the telescoping sum
for $a=1$.
