Compute the sum of binomial probability I have a sum that is $$\sum_{k=\lfloor{\frac{n}{2}+1}\rfloor}^n\binom{n}{k}p^n$$
I try $n=3$ and $n=4$, I got 
$$\sum_{k=\lfloor{\frac{3}{2}+1}\rfloor}^3\binom{n}{k}p^3=4p^3$$
$$\sum_{k=\lfloor{\frac{4}{2}+1}\rfloor}^4\binom{n}{k}p^4=5p^4$$
How do I do it for all $n$?
 A: For the sake of brevity let's define:
\begin{equation}
A = \sum_{k=\lfloor\frac{n}{2}\rfloor + 1}^n \binom{n}{k}p^n.
\end{equation}
First of all notice that by the Binomial Theorem we have:
\begin{equation}
2^n = (1+1)^n = \sum_{k=0}^n \binom{n}{k}
\end{equation}
and so multiplying by $p^n$ we get
\begin{equation}
(2p)^n = \sum_{k=0}^n \binom{n}{k}p^n.
\end{equation}
Notice that when $n$ is odd, the binomial coefficients $\binom{n}{0},\binom{n}{1},\dots,\binom{n}{n}$ are symmetric in the following sense:
\begin{equation}
\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\binom{n}{k} = 
\sum_{\lfloor\frac{n}{2}\rfloor + 1}^n\binom{n}{k}
\end{equation}
(think of Pascal's triangle: an example of an odd row is $1, 3, 3, 1$, and $1 + 3 = 3 + 1 = 4$). Thus, if $n$ is odd we have
\begin{equation}
\sum_{\lfloor\frac{n}{2}\rfloor+1}^n\binom{n}{k} = \frac{2^n}{2} = 2^{n-1}
\end{equation}
and hence $A = 2^{n-1}p^n$ whenever $n$ is odd. Can you see how to do the case where $n$ is even? 
A: Hint. Note that $\lfloor{\frac{n}{2}\pm 1}\rfloor=\lfloor{\frac{n}{2}}\rfloor\pm 1$, and $\binom{n}{k}=\binom{n}{n-k}$ for $k=0,\dots,n$. 
Hence
$$2^n=\sum_{k=0}^n\binom{n}{k}=\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor-1}\binom{n}{k}+\binom{n}{\lfloor{\frac{n}{2}}\rfloor}+\sum_{k=\lfloor{\frac{n}{2}}\rfloor+1}^n\binom{n}{k}\\
=\sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor-1}\binom{n}{n-k}+\binom{n}{\lfloor{\frac{n}{2}}\rfloor}+\sum_{k=\lfloor{\frac{n}{2}}\rfloor+1}^n\binom{n}{k}\\
\\
=\sum_{j=n-\lfloor{\frac{n}{2}}\rfloor+1}^{n}\binom{n}{j}+\binom{n}{\lfloor{\frac{n}{2}}\rfloor}+\sum_{k=\lfloor{\frac{n}{2}}\rfloor+1}^n\binom{n}{k}.$$
Can you take it from here?
