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  $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$
 A: An approach using the Lagrange inversion theorem: let $f(z)$ be analytic around $z=z_0$ with $f'(z_0)\neq0$; then $w=f(z)$ has an inverse $z=g(w)$ analytic around $w=f(z_0)$ with $$g^{(n)}\big(f(z_0)\big)=\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}\left(\frac{z-z_0}{f(z)-f(z_0)}\right)^n.\qquad(n>0)$$
We apply this to $f(z)=\dfrac{\sqrt{z}-1}{\sqrt{z}+1}$ and $z_0=1$, thus $g(w)=\left(\dfrac{1+w}{1-w}\right)^2$ and $$\lim_{z\to1}\frac{d^n}{dz^n}(1+\sqrt{z})^{2n+2}=g^{(n+1)}(0)=\color{blue}{4(n+1)!(n+1)}$$ since $g(w)=1+\displaystyle\frac{4w}{(1-w)^2}=1+4w\frac{d}{dw}\frac1{1-w}=1+4\sum_{n=1}^\infty nw^n$ for $|w|<1$.
A: 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.
A: A partial answer
Let
$$F=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}$$
So we can write $F=G+H$, where
$$G=\sum_{j=0}^{n+1}{2n+2\choose 2j}{j \choose n}, \quad H=\sum_{j=1}^{n+1} {2n+2 \choose 2j-1}{j-1/2 \choose n}$$
Only 2 terms in $G$ are nonzero, when $j=n,n+1$.
Hence $$G={2n+2 \choose 2n} {n \choose n}+{2n+2 \choose 2n+2}{n+1 \choose n}=2(n+1)^2.$$
Though Mathematica gives $H$ in terms of hypergemetric $_pF_q$, which gives $H=2(n+1)^2$ (numerically) again!
I wish to come back.
