<s>[INST] 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  $$n!\sum_{k=0}^{2n+2}\binom{2n+2}{k}\binom{\frac{k}{2}}{n}=n!\cdot4(n+1)^2?$$
 [/INST] 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.
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