Proving $\sum_{j=1}^{n+1} {2n+2 \choose 2j-1} {j-1/2 \choose n}=2(n+1)^2$ I have given a partial answer to a very interesting recent post Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$, where it remains to prove that
$$\sum_{j=1}^{n+1} {2n+2 \choose 2j-1} {j-1/2 \choose n}=2(n+1)^2.$$
Wolfarm Mathematica supports this result, I want your help in proving this.

We seek to prove that
$$\sum_{j=1}^{n+1} {2n+2\choose 2j-1} {j-1/2\choose n}
= 2(n+1)^2.$$
The LHS is
$$\sum_{j=0}^{n} {2n+2\choose 2j+1} {j+1/2\choose n}
= \sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n}
\frac{1-(-1)^j}{2}.$$
The first piece $A$ is
$$\frac{1}{2}
\sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n}
= \frac{1}{2} [w^n]
\sum_{j=0}^{2n+2} {2n+2\choose j} (1+w)^{j/2}
\\ = \frac{1}{2} [w^n] (1+\sqrt{1+w})^{2n+2}
= \frac{1}{2}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} (1+\sqrt{1+w})^{2n+2}.$$
Now put $1-\sqrt{1+w} = v$ so that $w=v(v-2)$ and $dw = 2(v-1) \; dv$
to obtain
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+1} (v-2)^{n+1}} (2-v)^{2n+2} (v-1)
\\ = \; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+1}} (v-2)^{n+1} (v-1)
\\ = {n+1\choose n-1} ((-1)\times 2)^2
- {n+1\choose n} ((-1)\times 2)
\\ = \frac{1}{2} (n+1) n \times 4
+ (n+1) \times 2 = 2(n+1)^2.$$
This is the claim, piece $B$ must be zero. And indeed we find
$$\frac{1}{2}
\sum_{j=0}^{2n+2} {2n+2\choose j} {j/2\choose n} (-1)^j
= \frac{1}{2}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}} (1-\sqrt{1+w})^{2n+2}.$$
Apply the same substitution to get
$$\; \underset{v}{\mathrm{res}} \;
\frac{1}{v^{n+1} (v-2)^{n+1}} v^{2n+2} (v-1)
= \; \underset{v}{\mathrm{res}} \;
\frac{1}{(v-2)^{n+1}} v^{n+1} (v-1) = 0$$
as desired.
