Solve Compound Sum for Unknown: a, c, l, and m

[ktoz]

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 = \frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}

where
a = 1
c = 2
l = 2
m = 4

30 = \frac{(m + l)!(a(l + 1) + cm)}{m!(l + 1)!}

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 = \frac{n(n + 1)}{2}
n = \frac{(sqrt(8v + 1) - 1)}{2}

Can the same sort of thing be done with the above compound sum formula?

 

[ktoz]

Omitting all the extraneous stuff from above, it becomes

v = \frac{(m + l)!(al + a + cm)}{m!(l + 1)!} <br />

Rearranging terms and expanding factorials I got this

v = \frac {(al + a + cm)(m + 1)(m + 2)(m + 3) ... \times (m + l)}{1 \times 2 \times 3 ... \times (l + 1)}

But what I don't know is if there is any way to rearrange the above to isolate the m term. Anyone know if there is a general way to do that?

 

[Office_Shredder]

Your notation is incredibly confusing... why do you need four variables to define an arithmetic sequence? At worst you should only need three

 

[ktoz]

Your notation is incredibly confusing...

Sorry. Highest level math I took in school was Algebra 2 (twenty years ago) so I'm kind of making it up as I go along.

Why do you need four variables to define an arithmetic sequence? At worst you should only need three

It's to allow for compound summation. for example

a = real = starting point
c = real = constant difference
m = natural = zero based term index
l = natural = zero based summation level

Example 1:
a = 1, c = 1, m = 5, l = 4

Sum table:
0 1 2 3 4 5 m
----------------------------------------
1 1 1 1 1 1 c
----------------------------------------
1 2 3 4 5 6 l = 0
1 3 6 10 15 21 l = 1
1 4 10 20 35 56 l = 2
1 5 15 35 70 126 l = 3
1 6 21 56 126 252 l = 4

Plugged into formula
252 = \frac{(5 + 4)!(1 \times 4 + 1 + 1 * 5)}{5!(4 + 1)!}

Example 2:
a = 1, c = 5, m = 5, l = 4

Sum table:
0 1 2 3 4 5 m
----------------------------------------
0 5 5 5 5 5 c
----------------------------------------
1 6 11 16 21 26 l = 0
1 7 18 34 55 81 l = 1
1 8 26 60 115 196 l = 2
1 9 35 95 210 406 l = 3
1 10 45 140 350 756 l = 4

Plugged into formula
756 = \frac{(5 + 4)!(1 \times 4 + 1 + 5 * 5)}{5!(4 + 1)!}

What the formula does isn't as much of a concern as how to isolate "m"