For a nonnegative integer $b$, and $|x|<1$, what is the function given by the power series $$ \sum_{n=0}^\infty \binom{b+2n}{b+n} x^n. $$
For $b=0$, this post shows $$ \sum_{n=0}^\infty \binom{2n}{n}x^n = (1-4x)^{-1/2}. $$
How do we proceed for an integer $b>0$?
I tried to start from the power series of $$ (1-x)^{-(b+2n)}, $$ but it doesn't work.
Note that the quoted identity is not difficult to verify using a variant of Lagrange Inversion. Introduce $$T(z) = w = \sqrt{1-4z}$$ so that $$z = \frac{1}{4} (1-w^2)$$ and $$dz = -\frac{1}{2} w \; dw$$
Then we seek to compute $$[z^n] \frac{2^b}{T(z)} (1+T(z))^{-b} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{2^b}{T(z)} (1+T(z))^{-b} \; dz.$$
Using the substitution this becomes $$- \frac{1}{2\pi i} \int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}} \frac{2^b}{w} (1+w)^{-b} \frac{1}{2} w \; dw \\ = - \frac{1}{2\pi i} \int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w)^{n+1} \times (1+w)^{n+1+b}} 2^{b-1} \; dw \\ = - \frac{1}{2\pi i} \int_{|w-1| = \epsilon} \frac{(-1)^{n+1} 4^{n+1}}{(w-1)^{n+1} \times (1+w)^{n+1+b}} 2^{b-1} \; dw.$$
It follows that the value of the integral is given by $$(-1)^n 4^{n+1} 2^{b-1} [(w-1)^n] \frac{1}{(1+w)^{n+1+b}} \\ = (-1)^n 4^{n+1} 2^{b-1} [(w-1)^n] \frac{1}{(2+(w-1))^{n+1+b}} \\ = (-1)^n 4^{n+1} 2^{b-1} \frac{1}{2^{n+1+b}} [(w-1)^n] \frac{1}{(1+(w-1)/2)^{n+1+b}} \\ = (-1)^n 2^{2n+b+1} \frac{1}{2^{n+1+b}} \frac{(-1)^n}{2^n} {n+n+b\choose n+b} \\= {2n+b\choose n+b}.$$
My suggestion : avoid binomial coefficient. Instead, write your series in its hypergeometric form : $$ \begin{eqnarray} \sum_{n=0}^\infty \binom{b+2n}{b+n} x^n &=& {}_2F_1\left( \left. \begin{array}{c} \frac 12 + \frac b2, 1 + \frac b2\\1+b\end{array}\right| 4x\right)\\ &=& \frac {2^b \left(\sqrt {1-4x} + 1\right)^{-b}}{\sqrt {1-4x}} \end{eqnarray} $$
This identity is well-known. You can find it on the Wolfram web site.
If you want to deduce your identity, you will have to apply twice the geometric sum $${}_1F_0\left( \left. \begin{array}{c} a\\-\end{array}\right| z\right) = \sum_{n=0}^{\infty} \frac{\left(a\right)_n z^n}{n!} = \left(1-z\right)^{-a}$$
where $(a)_n$ is the Pochhammer symbol.
Hope this helps.
