Recurrence formula for a series $x_n$ -- very stuck!!! The Problem:
$x_n$ is defined as $$\sum_{k=0}^\infty \frac{(n+k)!}{k!(2n+2k)!}(\frac{1}{2})^{2k} $$
and satisfies the recurrence: $x_n-2(2n+1)x_{n+1}=x_{n+2}$
But I cannot show that this is the case... 
My Attempt:
$$LHS=\sum_{k=0}^\infty\frac{(n+k)!}{k!(2n+2k)!}(\frac{1}{2})^{2k}-2(2n+1)\frac{(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{(n+k)!(2n+1+2k)(2n+2+2k)}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}-2(2n+1)\frac{(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+k)!(n+1+k)(2n+1+2k)-2(2n+1)(n+1+k)!}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+1+k)!( (2n+1+2k)-(2n+1))}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+1+k)!( 2k)}{k!(2n+2+2k)!}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{2(n+2+k)!( 2k)}{k!(2n+2+2k)!(n+2+k)}(\frac{1}{2})^{2k}$$
$$=\sum_{k=0}^\infty\frac{(n+2+k)!}{(k-1)!(2n+2+2k)!(2n+4+2k)}(\frac{1}{2})^{2(k-1)-1}$$
But from this point, I have no idea how to obtain $x_{n+2}$ from this...

My Solution:
To verify that the recurrence is true: we simply test the LHS and the RHS $\Rightarrow$
$$LHS=x_n - 2(2n+1)x_{n+1}$$
$$=\sum_{k=0}^\infty \frac{2(n+k+1)!( (2n+2k+1)-(2n+1) )}{k!(2n+2+2k)!}\times(1/2)^{2k}$$
Note that this is equal to $0$ when $k=0$ so we may just consider the sum starting at $k=1$
$$=\sum_{k=1}^\infty\frac{(n+k+1)!}{(k-1)!( 2(n+2)+2(k-1))!}\times(1/2)^{2(k-1)}$$
$$=\sum_{k=1}^\infty\frac{((n+2)+(k-1))!}{(k-1)!( 2(n+2)+2(k-1))!}\times(1/2)^{2(k-1)}$$
$$=\sum_{k=0}^\infty\frac{((n+2)+(k))!}{(k)!( 2(n+2)+2(k))!}\times(1/2)^{2(k)}$$
$$=x_{n+2} \text{ as required}$$
