closed form expression of a hypergeometric sum After playing around with transforms of a certain parametric integral, I am inclined to think that the linear combination $$f(n):=\dfrac1{n-2}\left({\,}_2F_1(\dfrac{n-2}{4n},\dfrac12;\dfrac{5n-2}{4n};-1)\right)+\dfrac1{n+2}\left({\,}_2F_1(\dfrac{n+2}{4n},\dfrac12;\dfrac{5n+2}{4n};-1)\right)$$ has a closed form for integer $n$. I know for example that $f(3)=\dfrac{1}{12^{3/4}}\dfrac{\Gamma(\frac14)^2}{\sqrt{\pi}}$. Any ideas?  
Edit: putting $a:=\frac14-\frac1{2n}$, we can define $g(a):= \frac{8}{1-4a}f(\frac{8}{1-4a})$ to get arguments closer to the "standard" notation used in formula collections. The question then becomes:

Which rational values of $a$, other than $a= \frac1{12}$, allow a closed form for$$g(a)=\dfrac1{a}\left({\,}_2F_1(a,\frac12;a+1;-1)\right)+\dfrac1{\frac{1}{2}-a}\left({\,}_2F_1(\dfrac{1}{2}-a,\frac12;\frac32-a;-1)\right)?$$

 A: If we use the original function,
$$f(n):=\tfrac1{n-2}\left({\,}_2F_1\big(\tfrac{n-2}{4n},\tfrac12;\tfrac{5n-2}{4n};-1\big)\right)+\tfrac1{n+2}\left({\,}_2F_1\big(\tfrac{n+2}{4n},\tfrac12;\tfrac{5n+2}{4n};-1\big)\right)$$
then there are several other $f(n)$ with a closed form. Let $K(k_d)$ be an elliptic integral singular value. Then,
$$f(3) = \sqrt{\frac{2}{3\sqrt3}}\; K(k_1)$$
$$f(4) = \sqrt{\frac{-1+\sqrt2}{2\sqrt2}}\; K(k_2)$$
$$f(5) = \frac{\sqrt2}{(\sqrt5\phi)^{5/4}}\; K(k_5)$$
$$f(6) = \sqrt{\frac{1}{12\sqrt3}}\; K(k_3)$$
$$f(12) = \sqrt{\alpha}\; K(k_6)$$
where $\phi$ is the golden ratio and $\alpha$ is a quartic root. 
P.S. Try as I might, I couldn't find anything else with $2<n<20$ which begs the question what makes these $n$ so special.
A: I have found two instances of $_2F_1(a,b;c;z), z=-1$ in
A. Erdelyi, Higher Transcendental Functions, Vol. 1 (and particularly, Sec. 2.8), Krieger Publishing.
that may be of help to you.
$$
_2F_1(a,b;1+a-b;-1)=2^{-a}\frac{\Gamma(1+a-b)\Gamma(1/2)}{\Gamma(1-b+a/2)\Gamma(1/2+a/2)}, \quad 1+a-b\ne0,-1,-2,...
$$
$$
(a+1)_2F_1(-a,1;b+2;-1)+(b+1)_2F_1(-b,1;a+2;-1)=2^{a+b+1}\frac{\Gamma(a+2)\Gamma(b+2)}{\Gamma(a+b+2)}, \quad a,b\ne-2,-3,-4,...
$$
And this one from the NIST Handbook of Mathematical Functions, which is a variation of the first one above
$$
_2F_1(a,b;1+a-b;-1)=\frac{\Gamma(1+a-b)\Gamma(a/2+1)}{\Gamma(1-b+a/2)\Gamma(1+a)}
$$
There was one additional relation for $_2F_1(a,b;c;z), z=-1$, but it was in terms of the digamma function.
