Sum of first m terms of a combinatorial number

[Quirinus Quirrell]

Dear Science Forums members,

I have a tricky problem that I hope one of you can help me with. (It's for a personal project, nothing to do with school.) I'm looking for a closed-form expression for the sum of the first through m-th terms of a combinatorial number. For those of you unfamiliar with combinatorial numbers, here's some useful reading: http://en.wikipedia.org/wiki/Combina..._number_system

Basically, the idea is this: for any non-negative integers k, and b, we can express the value of b as a sum of k terms of the form (r₁ choose k)+(r₂ choose k-1)+(r₃ choose k-2)...(r_k choose 1). For every t and s where t and s are non-negative integers such that t<s, it will be the case that r_t>r_s (this is just true by the definition of a combinatorial number).

For example, for k=5, we can express the number 36 as (7 choose 5)+(6 choose 4)+(2 choose 3)+(1 choose 2)+(0 choose 1).

Now, the sum of all five of these terms will be 36. But suppose I just want, say, the sum of the first two terms, four terms, or any arbitrary number of terms, and I don't want to exhaustively find every term and add all of them up. The question, then is this: given k, b, and m, where k is the total number of terms in the combinatorial number, b is the value of the combinatorial number, and m is the number of terms (starting with the first term) that we want to sum, what is the closed-form expression for the sum of those terms?

Admittedly, I am not certain that a closed-form expression even exists. If you can think of a reason why there might not be a closed-form expression for the above, please share it. In the eventuality that there is no closed-form expression, if you can think of a fast algorithm to find such a sum--something faster than just adding the terms individually--that would be helpful, too.

Edited January 8, 2013 by Quirinus Quirrell

[Time And Space]

sum of all binomial co-efficiants=2ⁿ