1. ## nested summation

Is there any way to calculate this efficiently?
$\displaystyle \sum_{i=min_N}^{max_N}\left\{ \sum_{j=min_{N-1}}^{i-1}\left\{ ...\left\{ \sum_{q=min_2}^{p-1} 1 ... \right\} \right\} \right\}$
It's only $\displaystyle N-1$ 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?
$\displaystyle \sum_{i=min_N}^{max_N}\left\{ \sum_{j=min_{N-1}}^{i-1}\left\{ ...\left\{ \sum_{q=min_2}^{p-1} 1 ... \right\} \right\} \right\}$
It's only $\displaystyle N-1$ nested summations where
the top limit of the next summation, is the iterator variable of the previous.
$\displaystyle \sum\limits_{i=a}^{b-1}\sum\limits_{j=a}^{i-1}f(j)=\sum\limits_{j=a}^{b-1}\sum\limits_{i=j+1}^{b-1}f(j)$
................_$\displaystyle =\sum\limits_{i=a}^{b-1}[b-(i+1)]f(i).$
$\displaystyle \sum\limits_{i_{n}=a}^{b-1}\sum\limits_{i_{n-1}=a}^{i_{n}-1}\cdots\sum\limits_{i_{1}=a}^{i_{2}-1}f(i_{1})=\sum\limits_{i=a}^{b-1}\frac{[b-(i+1)]^{(n-1)}}{(n-1)!}f(i),$
where $\displaystyle ~^{(\cdot)}$ stands for the factorial function, i.e. $\displaystyle n^{(k)}:=n(n-1)\cdots(n-k+1).$