Proving the Hypergeometric Sequence 
Question: How do you prove$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag{1}$$For $\Re(2x+2y+n+2)>0$

I'm not sure how to prove this. There are a multitude of other similar formulas. Some of which$$\begin{align*}_4F_3\left[\begin{array}{c c}\frac 12n+1,n,n,-x\\\frac 12n,x+n+1,1\end{array};-1\right] & =\dfrac {\Gamma(x+n+1)}{\Gamma(n+1)\Gamma(x+1)}\end{align*}\tag{2}$$For $\Re(2x-n+2)>0$$$_3F_2\left[\begin{array}{c c}\frac 12n+1,n,-x\\\frac 12n,x+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma\left(\frac 12n+\frac 12\right)}{\Gamma(n+1)\Gamma\left(x+\frac 12n+\frac 12\right)}\tag3$$
For $\Re(x)>-\frac 12$. I know that the general Hypergeometric Sequence can be written as$$_pF_q\left[\begin{array}{c c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_p\end{array};x\right]=\sum\limits_{k=0}^\infty\dfrac {(\alpha_1)_k(\alpha_2)_k\cdots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\cdots(\beta_p)_k}\dfrac {x^k}{k!}\tag4$$So $(1)$ becomes$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\sum\limits_{k=0}^{\infty}\dfrac {\left(\frac 12n+1\right)_k(n)_k(-x)_k(-y)_k}{\left(\frac 12n\right)_k(x+n+1)_k(y+n+1)_k}\dfrac {(-1)^k}{k!}\tag{5}$$However, how do you manipulate the RHS of $(5)$ into the RHS of $(1)$. I think that there could be a sort of elementary transformation involved, but if so, I'm not sure what.
 A: An established identity, useful to transform a $_4F_3 $ hypergeometric sequence with negative unit argument in a $_3F_2$ sequence, is 
$$_4F_3    \left[\begin{array}{c c}   a, b, c, d \\ a - b +1, a - c+1, a - d+1   \end{array};-1\right]  \\ =  \dfrac {\Gamma[a - b+1] \Gamma[a - c+1]}{\Gamma[a+1] \Gamma[a - b - c+1]} \,  _3F_2  \left[\begin{array}{c c} a/2 - d+1, b, c \\ a/2 + 1, a - d+1 \end{array};-1\right]$$
So, rewriting your series as 
$$_4F_3  \left[\begin{array}{c c} n, -x, -y, \frac 12n+1\\      n+x+1, n+y+1, \frac 12n \end{array};-1\right] \\$$
we obtain that it is equal to
$$\dfrac {\Gamma(n+x+1)\Gamma(n+y+1)}{\Gamma(n+1)\Gamma(n+x+y+1)} \\ _3F_2  \left[\begin{array}{c c} 0, -x, -y \\ n/2 + 1, n/2 \end{array};-1\right]   $$
and since the $_3F_2$ sequence, as a result of the zero among its terms, is equal to $1$, we get your identity.

The same method can be used to generate the second identity. Rewriting it as
$$\begin{align*}_4F_3\left[\begin{array}{c c} n, -x, n,\dfrac 12n+1 \\ n+x+1, 1, \dfrac 12n \end{array};-1\right] & \end{align*}$$
we get
$$  =\dfrac {\Gamma(n+x+1) \Gamma (1)}{\Gamma(n+1)\Gamma(x+1)} \, _3F_2  \left[\begin{array}{c c} 0, -x, n \\ n/2 + 1, n/2 \end{array};-1\right]   $$
where again the $_3F_2$ sequence is equal to $1$, giving your identity.

Lastly, another known identity can be used to express in closed form a $_3F_2$ hypergeometric function with negative unit argument:
$$_3F_2 \left[\begin{array}{c c} a, \dfrac a2+1,c \\ \dfrac a2,a-c+1 \end{array};-1 \right]=\dfrac {\sqrt {\pi} \,\, \Gamma(a-c+1)}{2^n \, \Gamma \left(\frac a2+1 \right)\Gamma \left(\frac{a+1}{2}-c \right)}$$
This can be used to prove the third identity of the OP. Rewriting it as follows we get
$$_3F_2 \left[\begin{array}{c c} n, \frac 12n+1,-x \\ \dfrac 12n,n+x+1 \end{array};-1 \right]=\dfrac {\sqrt {\pi} \,\, \Gamma(n+x+1)}{2^n \Gamma \left(\frac n2+1 \right) \Gamma \left(\frac{n+1}{2}+x \right)}$$ 
Now since it is known that, for given $z $,
$$\dfrac {\Gamma (z)}{\Gamma (2z)}= \dfrac {\sqrt {\pi}}{2^{2z-1}\, \Gamma \left(z+\frac{1}{2} \right)}$$
setting $z=(n+1)/2 \,\,$ we have
$$\dfrac {\Gamma (\frac 12n+ \frac 12 )}{\Gamma (n+1)}= \dfrac {\sqrt {\pi}}{2^{n} \Gamma \left(\frac {1}{2}n+1 \right)}$$
Substituting in the equation above we obtain
$$_3F_2 \left[\begin{array}{c c} n, \frac 12n+1,-x \\ \frac 12n,n+x+1 \end{array};-1 \right]=\dfrac {\Gamma(n+x+1) \Gamma \left(\frac 12n + \frac 12 \right)}{\Gamma(n+1) \Gamma \left(\frac 12n + x+ \frac 12  \right)}$$ 
