<s>[INST] Prove $\frac{1\cdot 3\cdot 5\cdots (2n - 1)}{ 2^n(n + 1)!}\cdot 4^n= \frac{1}{n+1} {2n\choose n}$ Prove: 
$$
 \frac{1\times 3\times 5\times \cdots \times (2n - 1)}{2^n (n + 1)!} \times 4^n = \frac{1}{n+1} \binom{2n}{n}
$$
-Sorry I don't know how to do choose notation in stack exchange.
I'm having a bit of trouble proving this, I tried to prove this by induction and would get stuck. Any help is appreciated in advance.   
 [/INST] By multiplying both numerator and denominator by $\;2\cdot 4\cdot 6 \cdot \ldots \cdot (2n)=2^nn!$ we get
\begin{align}
\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2^n(n+1)!}&=\frac{1\cdot \color{red}{2}\cdot 3\cdot \color{red}{4}\cdot 5\cdot \color{red}{6}\cdot \ldots \cdot (2n-1)\cdot\color{red}{2n}}{2^n(n+1)!\cdot \color{red}{2^nn!}}\\
&=\frac{1}{4^n(n+1)}\cdot\frac{(2n)!}{n!\cdot n!}\\
&=\color{blue}{\frac{1}{4^n(n+1)}{2n \choose n}}
\end{align}
Then
$$\boxed{\color{blue}{4^n\left[\frac{1\cdot 3\cdot 5\cdot \ldots \cdot (2n-1)}{2^n(n+1)!}\right]=\frac{1}{n+1}{2n \choose n}}}$$
</s>