Find $E[B 1_{\{ B\ge \frac{n}{2} \} }]$ where $B$ is Binomial $(n,\frac{1}{2})$. How to find the following expectation
$$E[B 1_{\{ B\ge \frac{n}{2} \} }]$$
where $B$ is Binomial random variable with  $(n,\frac{1}{2})$.
Here is what I did. The pmf  in this case is given by
\begin{align}
E[B 1_{\{ B\ge \frac{n}{2} \} }]=  \sum_{k \ge  \lceil n/2 \rceil }^n k  {n \choose k} \left( \frac{1}{2}\right)^n
\end{align}
However, I am not sure how to compute this sum. An upper bound would be fine too.
 A: Note that we have
$$
E[B 1_{\{ B\ge \frac{n}{2} \} }]=\frac n2 - E[B1_{\{B< \frac n2\}}]. 
$$
Then we evaluate $E[B1_{\{B< \frac n2\}}]$. We have
$$
E[B1_{\{B< \frac n2\}}]=\sum_{k<\frac n2} k\binom nk \frac 1{2^n}=\sum_{k<\frac n2} n \binom{n-1}{k-1} \frac1{2^n}. 
$$
The second equality is due to $k\binom nk = n \binom{n-1}{k-1}$.
Now, the index $k$ of the sum is up to $k\leq \frac n2 -1$ if $n$ is even,  and $k\leq \frac{n-1}2$ if $n$ is odd. Then $k-1\leq \frac n2-2$ if $n$ is even, and $k-1\leq \frac{n-3}2$ if $n$ is odd.
We apply the symmetry of binomial coefficients $\binom{n-1}k=\binom{n-1}{n-1-k}$.
Then the last sum is
$$
\begin{cases}
\frac n4 - \frac n{2^n}\binom{n-1}{\frac n2-1} &\mbox{if } n \mbox{ is even}\\
\frac n4- \frac12 \frac n{2^n}\binom{n-1}{(n-1)/2} &\mbox{if } n \mbox{ is odd}\\
\end{cases}.
$$
Therefore,
$$
E[B 1_{\{ B\ge \frac{n}{2} \} }]=\begin{cases}
\frac n4 + \frac n{2^n}\binom{n-1}{\frac n2-1} &\mbox{if } n \mbox{ is even}\\
\frac n4+ \frac12 \frac n{2^n}\binom{n-1}{(n-1)/2} &\mbox{if } n \mbox{ is odd}\\
\end{cases}.
$$
A: If $n$ is odd, it's $1/2$ due to symmetry.
If $n$ is even, it's
$$
\frac{1}{2}(1 - P(B = n/2)) + P(B = n/2)
$$
again due to symmetry.
Here, by symmetry I mean $P(B = k) = P( B = n-k)$ for any $k= 0, \dots, n$.
[EDIT] Seems that I misread the question; my apologies.
