- #1
ktoz
- 171
- 12
I came up with this formula for compound summation but don't know how to represent it correctly. Is there a standard way to represent an arbitrary level of compound summation where left sums depend on the sum directly to their right which depends on the sum to their right etc...?
level: 1
[tex]
\sum_{j=0}^m a + cj = \frac{(m + 1)(2a + cm)}{2}
[/tex]
level: 2
[tex]
\sum_{j=0}^m \frac{(j + 1)(2a + cj)}{2} = \frac{(m + 1)(m + 2)(3a + cm)}{6}\\
[/tex]
level: 3
[tex]
\sum_{j=0}^m \frac{(j + 1)(j + 2)(3a + cj)}{6} = \frac{(m + 1)(m + 2)(m + 3)(4a + cm)}{24}\\
[/tex]
level: 4
[tex]
\sum_{j=0}^m \frac{(j + 1)(j + 2)(j + 3)(4a + cj)}{24} = \frac{(m + 1)(m + 2)(m + 3)(m + 4)(5a + cm)}{120}\\
[/tex]
level: n
[tex]
= \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!}
[/tex]
For level "n" would something like this be correct?
[tex]
\sum_{j=0}^m \sum_{j=0}^m ... \sum_{j=0}^m a + cj = \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = a(n + 1) + cm \prod_{j=1}^n \frac{m + j}{j + 1}
[/tex]
Or is there a way to explicitly specify how many levels (n) of compound summation are desired?
Thanks
level: 1
[tex]
\sum_{j=0}^m a + cj = \frac{(m + 1)(2a + cm)}{2}
[/tex]
level: 2
[tex]
\sum_{j=0}^m \frac{(j + 1)(2a + cj)}{2} = \frac{(m + 1)(m + 2)(3a + cm)}{6}\\
[/tex]
level: 3
[tex]
\sum_{j=0}^m \frac{(j + 1)(j + 2)(3a + cj)}{6} = \frac{(m + 1)(m + 2)(m + 3)(4a + cm)}{24}\\
[/tex]
level: 4
[tex]
\sum_{j=0}^m \frac{(j + 1)(j + 2)(j + 3)(4a + cj)}{24} = \frac{(m + 1)(m + 2)(m + 3)(m + 4)(5a + cm)}{120}\\
[/tex]
level: n
[tex]
= \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!}
[/tex]
For level "n" would something like this be correct?
[tex]
\sum_{j=0}^m \sum_{j=0}^m ... \sum_{j=0}^m a + cj = \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = a(n + 1) + cm \prod_{j=1}^n \frac{m + j}{j + 1}
[/tex]
Or is there a way to explicitly specify how many levels (n) of compound summation are desired?
Thanks
Last edited: