Solve the equation: $\;$ $\dfrac{C^{2x}_{x+1}}{C^{2x+1}_{x-1}} = \dfrac{2}{3}$, $\;$ $x \in N$
$\dfrac{C^{2x}_{x+1}}{C^{2x+1}_{x-1}} = \dfrac{2}{3}$
i.e. $\;$ $\dfrac{2x!}{\left(x + 1\right)! \left(2x - x - 1\right)!} \div \dfrac{\left(2x + 1\right)!}{\left(x - 1\right)! \left(2x + 1 - x + 1\right)!} = \dfrac{2}{3}$
i.e. $\;$ $\dfrac{2x!}{\left(x + 1\right)! \left(x - 1\right)!} \times \dfrac{\left(x - 1\right)! \left(x + 2\right)!}{\left(2x + 1\right)!} = \dfrac{2}{3}$
i.e. $\;$ $\dfrac{2x!}{\left(x + 1\right)!} \times \dfrac{\left(x + 2\right) \left(x + 1\right)!}{\left(2x + 1\right) \left(2x\right)!} = \dfrac{2}{3}$
i.e. $\;$ $\dfrac{x + 2}{2x + 1} = \dfrac{2}{3}$
i.e. $\;$ $3x + 6 = 4x + 2$
i.e. $\;$ $x = 4$