What is the closed form for the sum of binomial coefficients over any interval?

[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.

 

[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

 

[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) = Σ_{k=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

 

[Svein]

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.

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

 

[Svein]

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

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