- #1
ktoz
- 171
- 12
Hi
I have a summation formula that can calculate the compound sum of terms
(a + c0) + (a + c1) + (a + c2) ... + (a + cm) to any level. Or put another way, it can sum the terms, sum the sums of the terms sum the sums of the sums of the terms etc.
Given
a = element of reals
c = element of reals
l = element of naturals
m = element of naturals
where
a = 1
c = 1
l = 1
m = 4
10 = [tex]\frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}[/tex]
where
a = 1
c = 2
l = 2
m = 4
30 = [tex]\frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}[/tex]
etc
What I'm wondering is, if given a compound sum and an unknown m, is it possible to do some sorcery and solve for m?
For example, with the simple case of summing naturals
v = [tex] \frac{n(n + 1)}{2}[/tex]
n = [tex]\frac{(sqrt(8v + 1) - 1)}{2}[/tex]
Can the same sort of thing be done with the above compound sum formula?
I have a summation formula that can calculate the compound sum of terms
(a + c0) + (a + c1) + (a + c2) ... + (a + cm) to any level. Or put another way, it can sum the terms, sum the sums of the terms sum the sums of the sums of the terms etc.
Given
a = element of reals
c = element of reals
l = element of naturals
m = element of naturals
where
a = 1
c = 1
l = 1
m = 4
10 = [tex]\frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}[/tex]
where
a = 1
c = 2
l = 2
m = 4
30 = [tex]\frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}[/tex]
etc
What I'm wondering is, if given a compound sum and an unknown m, is it possible to do some sorcery and solve for m?
For example, with the simple case of summing naturals
v = [tex] \frac{n(n + 1)}{2}[/tex]
n = [tex]\frac{(sqrt(8v + 1) - 1)}{2}[/tex]
Can the same sort of thing be done with the above compound sum formula?
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