I need to find a closed form expression in terms of $c$, $n$, $x$ and $y$ for
$$ \sum_{j=0}^{n}\rho^{c-j}\frac{x^j}{j!}\frac{y^{c-2j}}{\left(c-2j\right)!} $$
where $c$ and $\rho$ are just constants. I just can't spot the insight, not really experienced with finding closed forms for finite sums...perhaps this series is a particular type of finite series (not geometric, but maybe something else?).....
You can use the fact that $c=2n$ (though I doubt this helps).
On
To answer my own question using Mathematica, the closed form appears to be
$$ \frac{2^{c-1} (\rho y)^{c} (-\rho y^2)^{\frac{1}{2}-\frac{c}{2}}}{x^{\frac{1}{2}-\frac{c}{2}}c!}U(\frac{1-c}{2},\frac{3}{2},\frac{-\rho y^2}{4x}) $$
using $c=2n$, where $U(a,b,c)$ is the confluent hypergeometric function of the second kind.
Note that$$\sum_{k=0}^{n}\rho^{2n-k}\frac{x^{k}}{k!}\frac{y^{2n-2k}}{\left(2n-2k!\right)}=\left(y\rho\right)^{2n}\sum_{k=0}^{n}\frac{1}{k!\left(2n-2k\right)!}\left(\frac{x}{y^{2}\rho}\right)^{k}=$$ $$=\frac{\left(y\rho\right)^{2n}}{\left(2n\right)!}\sum_{k=0}^{n}\frac{\left(2n\right)!}{k!\left(2n-2k\right)!}\left(\frac{x}{y^{2}\rho}\right)^{k}=$$ $$=\frac{\left(y\rho\right)^{2n}}{2n!}\sum_{k=0}^{n}\frac{\left(2n\right)\left(2n-2\right)\cdots\left(2n-2k+2\right)\left(2n-1\right)\cdots\left(2n-2k+1\right)}{k!}\left(\frac{x}{y^{2}\rho}\right)^{k}=$$ $$=\frac{\left(y\rho\right)^{2n}}{2n!}\sum_{k=0}^{n}\frac{2^{2k}\left(n\left(n-1\right)\cdots\left(n-\left(k-1\right)\right)\left(n-\frac{1}{2}\right)\cdots\left(n-\frac{1}{2}-\left(k-1\right)\right)\right)}{k!}\left(\frac{y^{2}\rho}{x}\right)^{-k}=\frac{\left(y\rho\right)^{2n}}{2n!}\sum_{k=0}^{n}\frac{n\left(n-1\right)\cdots\left(n-\left(k-1\right)\right)\left(n-\frac{1}{2}\right)\cdots\left(n-\frac{1}{2}-\left(k-1\right)\right)}{k!}\left(\frac{y^{2}\rho}{4x}\right)^{-k}=$$ $$=\frac{\left(y\rho\right)^{2n}}{2n!}\sum_{k=0}^{\infty}\frac{n\left(n-1\right)\cdots\left(n-\left(k-1\right)\right)\left(n-\frac{1}{2}\right)\cdots\left(n-\frac{1}{2}-\left(k-1\right)\right)}{k!}\left(\frac{y^{2}\rho}{4x}\right)^{-k}$$ Recalling that the falling factorial is $$\left(a\right)_{k}=a\left(a-1\right)\cdots\left(a-\left(k-1\right)\right)$$ and its relation wih the Pochhammer symbol $\left(*\right)^{\left(k\right)}$ (or rising factorial)$$\left(a\right)_{k}=\left(-1\right)^{k}\left(-a\right)^{\left(k\right)}\,\,\,\,\,\,(1)$$ the sum is$$\frac{\left(y\rho\right)^{2n}}{2n!}\sum_{k=0}^{\infty}\frac{\left(n\right)_{k}\left(n+\frac{1}{2}-1\right)_{k}}{k!}\left(\frac{y^{2}\rho}{4x}\right)^{-k}=\frac{\left(y\rho\right)^{2n}}{2n!}\sum_{k=0}^{\infty}\frac{\left(-n\right)^{\left(k\right)}\left(-n-\frac{1}{2}+1\right)^{\left(k\right)}}{k!}\left(\frac{y^{2}\rho}{4x}\right)^{-k}=\frac{\left(-4x\rho\right)}{2n!}^{n}\left(-\frac{y^{2}\rho}{4x}\right)^{n} {}_2F_0\left(-n,1-n-\frac{1}{2};;\left(\frac{y^{2}\rho}{4x}\right)^{-1}\right)=\frac{\left(-4x\rho\right)}{2n!}^{n}U\left(-n,\frac{1}{2},-\frac{y^{2}\rho}{4x}\right)$$ where $_qF_p\left(a_{1},\dots,a_{p};b_{1},\dots,b_{q};x\right)$ is the generalized hypergeometric function and $U\left(a,b,z\right)$ is the confluent hypergeometric function of the second kind.