summation involving a hypergeometric 2F2 function im trying to find the closed form for the following
\begin{equation}
\sum_{n=0}^\infty \frac{c^n}{n!}\frac{(a)_n}{(b)_n}\frac{(\alpha+1/2)_n}{(\alpha+3/2)_n}{_2F_2}(-n,1-b-n;1-a-n,1/2;-\frac{d}{c})
\end{equation}
any help would be greatly appreciated. thank you
 A: $\sum\limits_{n=0}^\infty\dfrac{c^n}{n!}\dfrac{(a)_n}{(b)_n}\dfrac{\left(\alpha+\dfrac{1}{2}\right)_n}{\left(\alpha+\dfrac{3}{2}\right)_n}{_2F_2}\left(-n,1-b-n;1-a-n,\dfrac{1}{2};-\dfrac{d}{c}\right)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{\left(\alpha-n+\dfrac{3}{2}\right)(-a)^{(n)}(1-b-n)^{(k)}c^{n-k}d^k}{\left(\alpha+\dfrac{3}{2}\right)(-b)^{(n)}(1-a-n)^{(k)}\left(\dfrac{1}{2}\right)^{(k)}k!(n-k)!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-a)^{(n)}(n+b-1)_kc^{n-k}d^k}{(-b)^{(n)}(n+a-1)_k\left(\dfrac{1}{2}\right)^{(k)}k!(n-k)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{2n(-a)^{(n)}(n+b-1)_kc^{n-k}d^k}{(2\alpha+3)(-b)^{(n)}(n+a-1)_k\left(\dfrac{1}{2}\right)^{(k)}k!(n-k)!}$
(according to https://en.wikipedia.org/wiki/Falling_and_rising_factorials#Properties)
$=\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-a)^{(n)}\Gamma(n+b)\Gamma(n-k+a)c^{n-k}d^k}{(-b)^{(n)}\Gamma(n+a)\Gamma(n-k+b)\left(\dfrac{1}{2}\right)^{(k)}k!(n-k)!}-\sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{2n(-a)^{(n)}\Gamma(n+b)\Gamma(n-k+a)c^{n-k}d^k}{(2\alpha+3)(-b)^{(n)}\Gamma(n+a)\Gamma(n-k+b)\left(\dfrac{1}{2}\right)^{(k)}k!(n-k)!}$
(according to https://en.wikipedia.org/wiki/Falling_and_rising_factorials#Properties)
$=\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-a)^{(n+k)}\Gamma(n+k+b)\Gamma(n+a)c^nd^k}{(-b)^{(n+k)}\Gamma(n+k+a)\Gamma(n+b)\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{2(n+k)(-a)^{(n+k)}\Gamma(n+k+b)\Gamma(n+a)c^nd^k}{(2\alpha+3)(-b)^{(n+k)}\Gamma(n+k+a)\Gamma(n+b)\left(\dfrac{1}{2}\right)^{(k)}n!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(-b)^{(n+k)}(a)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{2n(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(2\alpha+3)(-b)^{(n+k)}(a)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{2k(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(2\alpha+3)(-b)^{(n+k)}(a)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(a)^{(n+k)}(-b)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^\infty\dfrac{2(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(2\alpha+3)(a)^{(n+k)}(-b)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}(n-1)!k!}-\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\dfrac{2(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(2\alpha+3)(a)^{(n+k)}(-b)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!(k-1)!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(a)^{(n+k)}(-b)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{2(-a)^{(n+k+1)}(b)^{(n+k+1)}(a)^{(n+1)}c^{n+1}d^k}{(2\alpha+3)(a)^{(n+k+1)}(-b)^{(n+k+1)}(b)^{(n+1)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{2(-a)^{(n+k+1)}(b)^{(n+k+1)}(a)^{(n)}c^nd^{k+1}}{(2\alpha+3)(a)^{(n+k+1)}(-b)^{(n+k+1)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k+1)}n!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-a)^{(n+k)}(b)^{(n+k)}(a)^{(n)}c^nd^k}{(a)^{(n+k)}(-b)^{(n+k)}(b)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{2a(1-a)^{(n+k)}(b+1)^{(n+k)}(a+1)^{(n)}c^{n+1}d^k}{(2\alpha+3)b(a+1)^{(n+k)}(1-b)^{(n+k)}(b+1)^{(n)}\left(\dfrac{1}{2}\right)^{(k)}n!k!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{4(1-a)^{(n+k)}(b+1)^{(n+k)}(a)^{(n)}c^nd^{k+1}}{(2\alpha+3)(a+1)^{(n+k)}(1-b)^{(n+k)}(b)^{(n)}\left(\dfrac{3}{2}\right)^{(k)}n!k!}$
$=\mathrm{F}^{2:1;0}_{2:1;1}\Bigg(\begin{matrix}-a,b&:&a&;&-&\\a,-b&:&b&;&\dfrac{1}{2}&\end{matrix}\Bigg|c,d\Bigg)-\dfrac{2ac}{(2\alpha+3)b}\mathrm{F}^{2:1;0}_{2:1;1}\Bigg(\begin{matrix}1-a,b+1&:&a+1&;&-&\\a+1,1-b&:&b+1&;&\dfrac{1}{2}&\end{matrix}\Bigg|c,d\Bigg)-\dfrac{4d}{2\alpha+3}\mathrm{F}^{2:1;0}_{2:1;1}\Bigg(\begin{matrix}1-a,b+1&:&a&;&-&\\a+1,1-b&:&b&;&\dfrac{3}{2}&\end{matrix}\Bigg|c,d\Bigg)$
