Say that we have this polynomial $(f + (1-f))^n$ where $f$ and $n$ are some positive real numbers, except that $f$ is a constant, but $n$ is a variable.
That term can be expanded using the binomial theorem: \begin{equation}\begin{split} (f+(1-f))^n &= \sum_{i = 1}^n {n \choose i} f^i (1-f)^{n-i}\\ &= 1 \end{split}\end{equation}
I want to find the sum the upper half terms of the full expansion, namely:
\begin{equation} \sum_{i = \text{int}(n/2)+1}^n {n \choose i} f^i (1-f)^{n-i} \end{equation}
I don't want the sum of the first half terms, namely:
\begin{equation} \sum_{i = 1}^{\text{int}(n/2)} {n \choose i} f^i (1-f)^{n-i} \end{equation}
But we can use the lower part that I don't want to rephrase the upper part that I want as follows: \begin{equation} 1-\sum_{i = 1}^{\text{int}(n/2)} {n \choose i} f^i (1-f)^{n-i} \end{equation}
More specifically, I wish to have a function $g(n)$ that, for any $n$, it spits out a number that is the sum of the upper half parts.
My struggle so far is in finding a close-form solution. I can't go algebra with this weird looking sum.
My attempt
Use this equation: \begin{equation} \Big(\sum_{i = \text{int}(n/2)+1}^n {n \choose i} f^i (1-f)^{n-i}\Big)+ \Big(\sum_{i = 1}^{\text{int}(n/2)} {n \choose i} f^i (1-f)^{n-i}\Big) = 1 \end{equation}
And rephrase it so that we end up having only $n$ on one side alone, and no $n$s on the other side.
And I am stuck here. Is this even approachable?
You can express it using hypergeometric functions. If $n = 2m$ is even, it is (according to Maple)
$$ {2\,m\choose m+1}{f}^{m+1} \left( 1-f \right) ^{m-1} {\mbox{$_2$F$_1$}(1,-m+1;\,m+2;\,{\frac {f}{-1+f}})} $$
while if $n=2m+1$ is odd,
$${2m+2 \choose m+1} \dfrac{{\mbox{$_2$F$_1$}(1,2\,m+2;\,m+2;\,f)}{f}^{m+1} \left( 1-f \right) ^{m+1}}{2} $$