# Binomial coefficients sum

1. Dec 12, 2015

### aaaa202

Is there a way to find the following sum in closed form:
∑K(N,n) , where K(N,n) is the binomial coefficient and the sum can extend over any interval from n=0..N. I.e. not necessarily n=0 to N in which case on can just use the binomial theorem.

2. Dec 12, 2015

### Buzz Bloom

Hi aaaa:

Technically speaking any specific finite sum is a closed form. So, I am assuming what you want is a close form for the sum when the range on the index variable is not fixed, but also a variable. If I am correct about this, I think you are out of luck. I don't think such a closed form exists.

However, the following Wikipedia article on the beta function may be of some help, although it is somewhat complicated. In particular, you may want to look at the discussion on
the "Cumulative distribution function". My vague memory is that the beta function is related to an approximation for the binomial distribution for large N.
https://en.wikipedia.org/wiki/Beta_distribution

Regards,
Buzz

3. Dec 12, 2015

### Buzz Bloom

Hi @aaaa202:

I had another thought about your problem. If what you want is a convenient way to calculate a sum of the terms in a binomial sequence,
S(N,i,j) = Σk=i to j K(N,k) ,​
then it is not too difficult to set up a spreadsheet to do this. Is this what you want?

Regards,
Buzz

4. Dec 12, 2015

### Svein

Well, we can find an upper limit: $2^{N}=(1+1)^{N}=\sum_{n=0}^{N}\begin{pmatrix} N\\ n\\ \end{pmatrix}1^{n}\cdot 1^{N-n}=\sum_{n=0}^{N}\begin{pmatrix} N\\ n\\ \end{pmatrix}$

5. Dec 13, 2015

### Svein

Some additional trivia: $\begin{pmatrix} N\\ n\\ \end{pmatrix}$ is symmetric ($\begin{pmatrix} N\\ n\\ \end{pmatrix}= \begin{pmatrix} N\\ N-n\\ \end{pmatrix}$). Therefore for odd N $\sum_{n=0}^{\frac{N+1}{2}}\begin{pmatrix} N\\ n\\ \end{pmatrix}=2^{N-1}$. For even N, there is a middle element.

Last edited: Dec 13, 2015