Solving mathematical summation

Question

How to solve the following sum:

$$\sum_{i = 0}^n \frac{(1/2^n)^i \cdot (1 - 1/2^n)^{n-i}}{i!\cdot(n-i)!}$$

Answer 1

After the rewrite of the question... $$ \sum_{i=0}^n\frac{1}{i!(n-i)!}\left(\frac{1}{2^n}\right)^i\left(1-\frac{1}{2^n}\right)^{n-i} $$ can be rewritten as $$ \frac{1}{n!}\sum_{i=0}^n\binom{n}{i}\left(\frac{1}{2^n}\right)^i\left(1-\frac{1}{2^n}\right)^{n-i}. $$ Then, by the binomial theorem, this is $$ \frac{1}{n!}\left[\frac{1}{2^n}+\left(1-\frac{1}{2^n}\right)\right]^n=\frac{1}{n!} $$