I am trying to simplify an expression involving summation as follows:
$\sum_{i=0}^{n-1} { {2n}\choose{i}}\cdot x^i$
where $n$ is an integer, and $x$ is a positive real number.
At a first glance, I can see that
$\sum_{i=0}^{2n} { {2n}\choose{i}}\cdot x^i = {(1+x)}^{2n}.$
But what if in the case when $i$ goes from 0 to $n-1$?