TITLE: Calculate $\sum_{i=1}^{[\frac{\sqrt n}{2}]}{n\choose i}$
QUESTION [5 upvotes]: It is known that $\sum_{i=1}^n {n \choose i}=2^n$. I am wondering what would be the sum if we change the upper limit to $\sqrt n/2$, i. e. How to calculate$$\sum_{i=1}^{[\frac{\sqrt n}{2}]}{n \choose i}?$$

REPLY [2 votes]: The terms in the sum grow rapidly, and for large $n$, the last term is much larger than all others.  The approximate answer is therefore $\binom{n}{\sqrt{n}/2}$ and the first $k$ terms of an asymptotic expansion can be had by adding the last $k$ terms in the sum.  Exact formulas to simplify the sum do not exist.