<s>[INST] Generating function for binomial coefficients $\binom{2n+k}{n}$ with fixed $k$ Prove that
$$
\frac{1}{\sqrt{1-4t}} \left(\frac{1-\sqrt{1-4t}}{2t}\right)^k = \sum\limits_{n=0}^{\infty}\binom{2n+k}{n}t^n,
\quad
\forall k\in\mathbb{N}.
$$
I tried already by induction over $k$ but i have problems showing the statement holds for $k=0$ or $k=1$.
 [/INST] This can be shown 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
$$[z^n] \frac{1}{T(z)} \left(\frac{1-T(z)}{2z}\right)^k
= \frac{1}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{1}{T(z)} \left(\frac{1-T(z)}{2z}\right)^k dz.$$
Using the substitution this becomes
$$- \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}}
\frac{1}{w} \left(\frac{1-w}{1/2(1-w^2)}\right)^k
\frac{1}{2} w \; dw
\\ = - \frac{1}{2} \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{4^{n+1}}{(1-w^2)^{n+1}}
\left(\frac{2}{1+w}\right)^k dw
\\ = -  \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{1}{(1-w)^{n+1}}
\frac{2^{2n+k+1}}{(1+w)^{n+k+1}} dw
\\ = (-1)^n  \frac{1}{2\pi i}
\int_{|w-1| = \epsilon} \frac{1}{(w-1)^{n+1}}
\frac{2^{2n+k+1}}{(1+w)^{n+k+1}} dw.$$
Now expand the second fraction about $w=1$ to get
$$\frac{1}{(1+w)^{n+k+1}}
= \frac{1}{(2+(w-1))^{n+k+1}}
= \frac{1}{2^{n+k+1}} \frac{1}{(1+1/2(w-1))^{n+k+1}}
\\ = \frac{1}{2^{n+k+1}}
\sum_{q\ge 0} {q+n+k\choose q} (-1)^q \frac{(w-1)^q}{2^q}.$$
We need $q=n$ for the residue in the integral, getting the final answer
$$(-1)^n \times 2^{2n+k+1} \times
\frac{1}{2^{n+k+1}}
{2n+k\choose n} (-1)^n \frac{1}{2^n}
= {2n+k\choose n}.$$
This MSE link has another calculation that is quite similar.
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