Calculating $\sum_{k=0}^{n} {n\choose k} \sqrt{k}$ I am trying to calculate the following expression but not sure how this can be solved yet.
$$\sum_{k=0}^{n} {n\choose k} \sqrt{k}$$
The wiki page (https://en.wikipedia.org/wiki/Binomial_coefficient) contains a similar expression with $\sqrt{k}$ replaced by $k$ or $k^2$. If the above expression can be solved, I also wonder if we can solve it for any $k^j$ with any real number $j>0$?

Let $X$ be a binomial random variable, with $n$ trials and probability $1/2$.  Your expression is exactly $2^nE[X^j]$. Using the central limit theorem, for large $n$, $X$ is well approximated by $\frac{\sqrt{n}}2Z+\frac{n}2$, where $Z$ is a standard normal random variable. I will then use the approximation
\begin{align}
E[X^j]
&\approx E\left[\left(\frac{\sqrt{n}}2Z+\frac{n}2\right)^j\right]\tag 1
\end{align}
to get an approximation for $E[X^j]$. I am not at this time able to quantify the error of the approximation in $(1)$, which is necessary to determine how good of an approximation the final answer is.
\begin{align}
2^nE[X^j]
  &\approx 2^nE\left[\left(\frac{\sqrt{n}}2Z+\frac{n}2\right)^j\right]
\\&            =2^{n}(n/2)^jE\left[(1+Z/\sqrt{n})^j\right]
\\&\stackrel{1}=2^{n}(n/2)^j\sum_{i=0}^\infty \binom{j}{i}E[Z^i]n^{-i/2}
\\&\stackrel{2}=\boxed{2^{n}(n/2)^j\sum_{i=0}^\infty \binom{j}{2i}(2i-1)!!\cdot n^{-i}}
\end{align}

*

*In $1$, we use the Taylor series for $f(x)=(1+x)^j=\sum_{i\ge 0}\binom{j}ix^i$, where the binomial coefficient is defined for non-integer $j$ as $$\binom{j}i=\frac{j(j-1)\cdots(j-i+1)}{i!}.$$

*In $2$, we reindex $i\gets 2i$, and use the known moments of the standard normal random variable.

We can then further approximate by only taking the first several terms of the infinite summation. For example, when $j=1/2$, you get
$$
\sum_{k=0}^n \binom{n}kk^{1/2}=2^nE[X^{1/2}]\approx 2^n(n/2)^{1/2}\left(1-\frac18n^{-1}-\frac{15}{128}n^{-2}-\dots\right),
$$
so that
$$
\sum_{k=0}^n \binom{n}kk^{1/2}\approx 2^n(n/2)^{1/2}(1+O(1/n))
$$
Again, use this with caution, because the goodness of this approximation depends on the the quality of the approximation $(1)$, which I am not able to quantify.
