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 $k=1$.

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$\ds{{1 \over \root{1 - 4t}}\,\pars{1 - \root{1-4t} \over 2t}^{k}
     =\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}\,,\quad
     \forall k\ \in\ \mathbb{N}}$

With $\ds{0\ <\ t\ <\ {1 \over 4}}$:
  \begin{align}&\color{#c00000}{\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}}
=\sum_{n = 0}^{\infty}\bracks{\oint_{\verts{z}\ =\ 1^{+}}{\pars{1 + z}^{2n + k} \over z^{n + 1}}%
\,{\dd z \over 2\pi\ic}}t^{n}
\\[5mm]&=\oint_{\verts{z}\ =\ 1^{+}}{\pars{1 + z}^{k} \over z}\sum_{n = 0}^{\infty}
\bracks{\pars{1 + z}^{2}\,t \over z}^{n}\,{\dd z \over 2\pi\ic}
=\oint_{\verts{z}\ =\ 1^{+}}{\pars{1 + z}^{k} \over z}
{1 \over 1 - \pars{1 + z}^{2}\,t/z}\,{\dd z \over 2\pi\ic}
\\[5mm]&=-\,{1 \over t}\oint_{\verts{z}\ =\ 1^{+}}
{\pars{1 + z}^{k} \over z^{2} + \pars{2 - 1/t}z + 1}
\,{\dd z \over 2\pi\ic}
=-\,{1 \over t}\oint_{\verts{z}\ =\ 1^{+}}
{\pars{1 + z}^{k} \over \pars{z - z_{-}}\pars{z - z_{+}}}\,{\dd z \over 2\pi\ic}
\end{align}

where $\ds{z_{\pm}}$ are the roots of
$\ds{z^{2} + \pars{2 - {1 \over t}}z + 1 = 0}$:
$$
z_{\pm} = {1 \pm \root{1 - 4t} - 2t \over 2t}
$$
such that $\ds{z_{+} > 1}$ and $\ds{\verts{z_{-}} < 1}$.

Then,
  \begin{align}&\color{#c00000}{\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}}
=-\,{1 \over t}\,{\pars{1 + z_{-}}^{k} \over z_{-} - z_{+}}
=-\,{1 \over t}\pars{-\,{t \over \root{1 - 4t}}}
\pars{1 + {1 - \root{1 - 4t} - 2t\over 2t}}^{k}
\end{align}

which simplifies to
$$\color{#66f}{\large%
{1 \over \root{1 - 4t}}\,\pars{1 - \root{1-4t} \over 2t}^{k}
     =\sum_{n = 0}^{\infty}{2n + k \choose n}t^{n}}
$$
