# Math Help - nested summation

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.
Thanx in advance

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.
Thanx in advance
I hope you are asking the following formula (this is what my signature reads)
$\sum\limits_{i=a}^{b-1}\sum\limits_{j=a}^{i-1}f(j)=\sum\limits_{j=a}^{b-1}\sum\limits_{i=j+1}^{b-1}f(j)$
................_ $=\sum\limits_{i=a}^{b-1}[b-(i+1)]f(i).$
You may easiliy generalize this as follows:
$\sum\limits_{i_{n}=a}^{b-1}\sum\limits_{i_{n-1}=a}^{i_{n}-1}\cdots\sum\limits_{i_{1}=a}^{i_{2}-1}f(i_{1})=\sum\limits_{i=a}^{b-1}\frac{[b-(i+1)]^{(n-1)}}{(n-1)!}f(i),$
where $~^{(\cdot)}$ stands for the factorial function, i.e. $n^{(k)}:=n(n-1)\cdots(n-k+1).$
You may also refer to discrete Taylor's formula.