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 $n$?
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?
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?
