1. ## nested summation

Is there any way to calculate this efficiently?
$
\sum_{i=min_N}^{max_N}\left\{
\sum_{j=min_{N-1}}^{i-1}\left\{
...\left\{
\sum_{q=min_2}^{p-1} 1
...
\right\}
\right\}
\right\}
$

It's only $N-1$ nested summations where
the top limit of the next summation, is the iterator variable of the previous.

2. Originally Posted by phsoftnet
Is there any way to calculate this efficiently?
$
\sum_{i=min_N}^{max_N}\left\{
\sum_{j=min_{N-1}}^{i-1}\left\{
...\left\{
\sum_{q=min_2}^{p-1} 1
...
\right\}
\right\}
\right\}
$

It's only $N-1$ nested summations where
the top limit of the next summation, is the iterator variable of the previous.
$\sum\limits_{i=a}^{b-1}\sum\limits_{j=a}^{i-1}f(j)=\sum\limits_{j=a}^{b-1}\sum\limits_{i=j+1}^{b-1}f(j)$
................_ $=\sum\limits_{i=a}^{b-1}[b-(i+1)]f(i).$
$\sum\limits_{i_{n}=a}^{b-1}\sum\limits_{i_{n-1}=a}^{i_{n}-1}\cdots\sum\limits_{i_{1}=a}^{i_{2}-1}f(i_{1})=\sum\limits_{i=a}^{b-1}\frac{[b-(i+1)]^{(n-1)}}{(n-1)!}f(i),$
where $~^{(\cdot)}$ stands for the factorial function, i.e. $n^{(k)}:=n(n-1)\cdots(n-k+1).$