Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ 
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .

Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^{\infty}\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{j}(x-1)^j. \end{align*}
Hence \begin{align*} f^{(n)}(1)&=n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2. \end{align*}
Is it correct? How to compute  $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$

This is just a supplement to the nice answer of @Cathedral. Here we close a gap and show
\begin{align*}
\color{blue}{\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{2r+1}=\frac{(2r)!!}{(2r+1)!!}}\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{r=0}^n}&\color{blue}{(-1)^r\binom{n}{r}\frac{1}{2r+1}}\\
&=\frac{1}{2}(-1)^nn!\sum_{r=0}^n\frac{1}{r+\frac{1}{2}}
\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=r+1}^n\frac{1}{r-q}\tag{2.1}\\
&=\frac{1}{2}(-1)^nn!\sum_{r=0}^n\underbrace{\operatorname{res}_{z=r}\left(\frac{1}{z+\frac{1}{2}}
\prod_{q=0}^{n}\frac{1}{z-q}\right)}_{f(z)}\tag{2.2}\\
&=\frac{1}{2}(-1)^nn!\left(-\operatorname{res}_{z=-\frac{1}{2}}f(z)
-\operatorname{res}_{z=-\infty}f(z)\right)\tag{2.3}\\
&=\frac{1}{2}(-1)^nn!\left(-\lim_{z=-\frac{1}{2}}\prod_{q=0}^n\frac{1}{z-q}
+\underbrace{\operatorname{res}_{z=0}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)}_{=0}\right)\tag{2.4}\\
&=\frac{1}{2}(-1)^{n+1}n!\prod_{q=0}^n\frac{1}{-\frac{1}{2}-q}\\
&=2^nn!\prod_{q=0}^n\frac{1}{2q+1}\\
&=\frac{2^nn!}{(2n+1)!!}\\
&\,\,\color{blue}{=\frac{(2n)!!}{(2n+1)!!}}
\end{align*}
and the claim (1) follows. Similarly we can show
\begin{align*}
\color{blue}{\sum_{r=0}^n(-1)^r\binom{n}{r}\frac{1}{2r-1}=-\frac{(2n)!!}{(2n-1)!!}}
\end{align*}

Comment:

*

*In (2.1) we use
\begin{align*}
\binom{n}{r}&=\frac{n!}{r!(n-r)!}\\
&=n!\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=0}^{n-r-1}\frac{1}{n-r-q}\\
&=n!\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=0}^{n-r-1}\frac{1}{q+1}\tag{$q\to\ n-r-1-q$}\\
&=n!(-1)^{n-r}\prod_{q=0}^{r-1}\frac{1}{r-q}\prod_{q=r+1}^n\frac{1}{r-q}
\end{align*}


*In (2.2) we write the summands as residue of a meromorphic function at the pole $z=q$.


*In (2.3) we use the sum of the residues of a meromorphic function at the poles $z=q, 0\leq q\leq r$ and $z=-\frac{1}{2}$ plus the residue at $\infty$ sum up to zero. This way we get rid of the sum and what is left are just two residues, the one at $z=-\frac{1}{2}$ and the one at $z=\infty$.


*In (2.4) we use the identity
\begin{align*}
\operatorname{res}_{z=\infty}f(z)=\operatorname{res}_{z=0}\left(-\frac{1}{z^2}f\left(\frac{1}{z}\right)\right)
\end{align*}
which transforms a residue at infinity to a residue at zero. We then find by inspection the residue of $f(z)$ at $z=\infty$ is zero.
