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 $k=1$.

Due to a recent comment on my other answer, I took a second look at this question and tried to apply a double generating function.
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=-n}^\infty\binom{2n+k}{n}x^ny^k\\
&=\sum_{n=0}^\infty\sum_{k=n}^\infty\binom{k}{n}\frac{x^n}{y^{2n}}y^k\\
&=\sum_{n=0}^\infty\frac{x^n}{y^{2n}}\frac{y^n}{(1-y)^{n+1}}\\
&=\frac1{1-y}\frac1{1-\frac{x}{y(1-y)}}\\
&=\frac{y}{y(1-y)-x}\\
&=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}-\frac{1-\sqrt{1-4x}}{1-\sqrt{1-4x}-2y}\right)\\
&=\frac1{\sqrt{1-4x}}\left(\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}+\color{#C00000}{\frac{2x/y}{1+\sqrt{1-4x}-2x/y}}\right)\tag{1}
\end{align}
$$
The term in red contains those terms with negative powers of $y$. Eliminating those terms yields
$$
\begin{align}
\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2n+k}{n}x^ny^k
&=\frac1{\sqrt{1-4x}}\frac{1+\sqrt{1-4x}}{1+\sqrt{1-4x}-2y}\\
&=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{2y}{1+\sqrt{1-4x}}\right)^k\\
&=\frac1{\sqrt{1-4x}}\sum_{k=0}^\infty\left(\frac{1-\sqrt{1-4x}}{2x}\right)^ky^k\tag{2}
\end{align}
$$
Equating identical powers of $y$ in $(2)$ shows that
$$
\sum_{n=0}^\infty\binom{2n+k}{n}x^n=\frac1{\sqrt{1-4x}}\left(\frac{1-\sqrt{1-4x}}{2x}\right)^k\tag{3}
$$
