I'm interested in finding
$$\sum_{k=0}^m \binom{n}{k}, \quad m<n$$
which form rows of Pascal's triangle. Surely $\sum\limits_{k=0}^n \binom{k}{m}$ using addition formula, but the one above involves hypergeometric functions and I don't know how to approach it.
EDIT: if possible, please don't solve it, just a few hints will do.
Best Answer
There really isn't a closed-form expression for the partial row sums of Pascal's triangle. The expression I imagine you're getting, $$\sum_{k=0}^m \binom{n}{k} = 2^n - \binom{n}{m+1} {}_2F_1(1,m+1-n;m+2;-1),$$ isn't really a closed-form; it's just the original sum expressed differently. (One of the answers to the MO question mentioned above by anon says this as well.)
So, why is this?
(In what follows, the original version contained a partial explanation, as the OP merely wanted hints. Now that the OP has completed the derivation for him/herself, I'm giving the full argument.)
We have $$ \binom{n}{m+1} {}_2F_1(1,m+1-n;m+2;-1) = \binom{n}{m+1} \sum_{k=0}^{\infty} \frac{1^{\overline{k}} (m+1-n)^{\overline{k}} (-1)^k}{(m+2)^{\overline{k}} k!}$$ $$= \binom{n}{m+1} \left[ 1 - \frac{1(m+1-n)}{(m+2)1} + \frac{2!(m+1-n)(m+2-n)}{(m+2)(m+3)2!} \mp \cdots \right]$$ $$= \frac{n!}{(m+1)!(n-m-1)!} \left[ 1 + \frac{n-m-1}{m+2} + \frac{(n-m-1)(n-m-2)}{(m+2)(m+3)} + \cdots \right]$$ $$= \binom{n}{m+1} + \binom{n}{m+2} + \binom{n}{m+3} + \cdots $$ $$= \sum_{k=m+1}^n \binom{n}{k}.$$
Thus $$\sum_{k=0}^m \binom{n}{k} = 2^n - \binom{n}{m+1} {}_2F_1(1,m+1-n;m+2;-1)$$ is just a rewrite of the original sum (together with the fact that $\sum_{k=0}^n\binom{n}{k} = 2^n$).
By the way, Concrete Mathematics spends some time discussing $\sum_{k=0}^m \binom{n}{k}$. On p. 165 they say
Then on p. 228 they discuss the sum again, in the context of Gosper's algorithm for finding partial hypergeometric sums. This gives an explanation for why there is no closed-form, as Gosper's algorithm either finds one or proves that no such one exists.
However, Exercise 9.42 on p. 492 asks to prove the asymptotic formula $$\sum_{k \leq \alpha n} \binom{n}{k} = 2^{n H(\alpha) - \frac{1}{2} \log_2 n + O(1)},$$ where $H(\alpha) = \alpha \log_2 \frac{1}{\alpha} + (1-\alpha) \log_2 (\frac{1}{1-\alpha})$, for $\alpha \leq \frac{1}{2}$.
There is an outline of the proof of this result in the answer section, so you can look that up if you want to see how they obtain it. See also robjohn's comments below.
If $\alpha \geq \frac{1}{2}$, then $$\sum_{k \leq \alpha n} \binom{n}{k} = 2^{n + O(1)},$$ since the sum is bounded above by $2^n$ and below by $2^{n-1}$.
(Page numbers are for the second edition of Concrete Mathematics.)