Help with writing compound sum

[ktoz]

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
$$\sum_{j=0}^m a + cj = \frac{(m + 1)(2a + cm)}{2}$$

level: 2
$$\sum_{j=0}^m \frac{(j + 1)(2a + cj)}{2} = \frac{(m + 1)(m + 2)(3a + cm)}{6}\\$$

level: 3
$$\sum_{j=0}^m \frac{(j + 1)(j + 2)(3a + cj)}{6} = \frac{(m + 1)(m + 2)(m + 3)(4a + cm)}{24}\\$$

level: 4
$$\sum_{j=0}^m \frac{(j + 1)(j + 2)(j + 3)(4a + cj)}{24} = \frac{(m + 1)(m + 2)(m + 3)(m + 4)(5a + cm)}{120}\\$$

level: n
$$= \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!}$$

For level "n" would something like this be correct?
$$\sum_{j=0}^m \sum_{j=0}^m ... \sum_{j=0}^m a + cj = \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = a(n + 1) + cm \prod_{j=1}^n \frac{m + j}{j + 1}$$

Or is there a way to explicitly specify how many levels (n) of compound summation are desired?

Thanks

 

[benorin]

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
$$\sum_{j=0}^m a + cj = \frac{(m + 1)(2a + cm)}{2}$$

level: 2
$$\sum_{j=0}^m \frac{(j + 1)(2a + cj)}{2} = \frac{(m + 1)(m + 2)(3a + cm)}{6}\\$$

level: 3
$$\sum_{j=0}^m \frac{(j + 1)(j + 2)(3a + cj)}{6} = \frac{(m + 1)(m + 2)(m + 3)(4a + cm)}{24}\\$$

level: 4
$$\sum_{j=0}^m \frac{(j + 1)(j + 2)(j + 3)(4a + cj)}{24} = \frac{(m + 1)(m + 2)(m + 3)(m + 4)(5a + cm)}{120}\\$$

level: n
$$= \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!}$$

For level "n" would something like this be correct?

$$\sum_{j=0}^m \sum_{j=0}^m ... \sum_{j=0}^m a + cj = \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = a(n + 1) + cm \prod_{j=1}^n \frac{m + j}{j + 1}$$

Or is there a way to explicitly specify how many levels (n) of compound summation are desired?

Thanks

You might try subscript your indicies like this:

$$\sum_{j_{n}=0}^m \sum_{j_{n-1}=0}^m \cdot\cdot\cdot \sum_{j_{1}=0}^{m} a + cj = \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = a(n + 1) + cm \prod_{j=1}^n \frac{m + j}{j + 1}$$

But that isn't right.

I think there is, but there is one problem (with my sum and with yours), namely this: that, in going from one level to the next, you replace $$m$$ by $$j$$ : a constant with an index.

So somthing like this would be more fitting:

$$\sum_{j_{n}=0}^m \sum_{j_{n-1}=0}^{j_{n}} \cdot\cdot\cdot \sum_{j_{0}=0}^{j_{1}} a + cj = \frac{(m + n)!(a(n + 1) + cm)}{m!(n + 1)!} = a(n + 1) + cm \prod_{j=1}^n \frac{m + j}{j + 1}$$
but that isn't right either.

You can't sum over a constant, and you can't fix (i.e. make constant) an index, so hmmm... I don't know exactly. I hope I gave you something to work with, even if it was only to illustrate the nature of the problem.