<s>[INST] How to write $\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}$ in closed form or short form? How to write $\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}$ in closed form or short form?
We can write the above sum as 
$\binom{0}{0}x+\binom{2}{1}x^2+\binom{4}{2}x^3+\binom{6}{3}x^4+\cdots+\binom{2N-2}{N-1}x^N$.
There is a formula $(1+x)^n=\sum_{k=0}^{n} \binom{n}{k}x^k$. 
But how to apply it?
Is there any short form or any function that would give short form ?
 [/INST] We can transform using the duplication formula DLMF
\begin{align}
\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}&=x\sum_{n=0}^{N-1} \frac{x^n}{n!}\frac{\Gamma(2n+1)}{\Gamma(n+1)}\\
&=\frac{x}{\sqrt{\pi}}\sum_{n=0}^{N-1} \frac{(4x)^n}{n!}\frac{\Gamma(n+1)\Gamma(n+1/2)}{\Gamma(n+1)}\\
&=\frac{x}{2\sqrt{\pi}}\sum_{n=0}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)-\frac{x}{\sqrt{\pi}}\sum_{n=N}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)\\
&=\frac{x}{\sqrt{1-4x}}-\frac{x}{\sqrt{\pi}}\sum_{n=N}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)
\end{align}
from the Newton's generalized binomial theorem. Now, shifting the index in the sum,
\begin{align}
\sum_{n=N}^{\infty} \frac{(4x)^n}{n!}\Gamma(n+1/2)&=\left( 4x \right)^N\sum_{k=0}^{\infty} \frac{(4x)^k}{(N+k)!}\Gamma(N+k+1/2)\\
&=\left( 4x \right)^N\sum_{k=0}^{\infty} \frac{(4x)^k}{k!}\frac{\Gamma(k+1)\Gamma(k+N+1/2)}{\Gamma(k+N+1)}\\
&=\left( 4x \right)^N\frac{\Gamma(N+1)}{\Gamma(N+1/2)}\,_2F_1\left( 1,N+1/2;N+1;4x) \right)
\end{align}
from the definition of the hypergeometric function. Finally, as noted by @Sil in the comment
\begin{equation}
\sum_{n=0}^{N-1} \binom{2n}{n}x^{n+1}=\frac{x}{\sqrt{1-4x}}- \binom{2N}{N}x^{N+1}\,_2F_1\left( 1,N+1/2;N+1;4x) \right)
\end{equation} 
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