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 $k=1$.

We proceed by induction on $k$.
For $k = 0$, we have $\displaystyle \sum\limits_{m=0}^{\infty} \binom{2m}{m}t^m = \frac{1}{\sqrt{1-4t}}$ 
and for $k = 1$, we have 
\begin{align*} \sum\limits_{m=0}^{\infty} \binom{2m+1}{m+1}t^m &= \sum\limits_{m=0}^{\infty} \left(2 - \frac{1}{m+1}\right)\binom{2m}{m}t^m \\&= \frac{2}{\sqrt{1-4t}} - C(t)\\&= \frac{2}{\sqrt{1-4t}} -\frac{2}{1+\sqrt{1-4t}} \\&= \frac{2}{\sqrt{1-4t}(1+\sqrt{1-4t})}\end{align*}
Where, $\displaystyle C(t) = \sum\limits_{m=0}^{\infty} \frac{1}{m+1}\binom{2m}{m}z^m = \frac{2}{1+\sqrt{1-4t}}$ is the generating function of Catalan Numbers.
Assuming the result holds for $k$ we prove for $k+1$:
\begin{align*}\sum\limits_{m=0}^{\infty} \binom{2m+k+1}{m+k+1}t^m &= \sum\limits_{m=0}^{\infty} \left(2 - \frac{k+1}{m+k+1}\right)\binom{2m+k}{m+k}t^m\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{k+1}{t^{k+1}}\int_0^t \sum\limits_{m=0}^{\infty} \binom{2m+k}{m+k}t^{m+k}\,\mathrm{d}t\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{2^k(k+1)}{t^{k+1}}\int_0^t \frac{t^k(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}}\,\mathrm{d}t\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - \frac{k+1}{2^kt^{k+1}}\int_0^t \frac{(1-\sqrt{1-4t})^{k}}{\sqrt{1-4t}}\,\mathrm{d}t\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-p}}{\sqrt{1-4t}} - \frac{k+1}{2^{k+1}t^{k+1}} \frac{(1-\sqrt{1-4t})^{k+1}}{k+1}\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-k}}{\sqrt{1-4t}} - 2^{k+1}(1+\sqrt{1-4t})^{-(k+1)}\\
&= \frac{2^{k+1}(1+\sqrt{1-4t})^{-(k+1)}}{\sqrt{1-4t}}\\
\end{align*}
