Question

Solve the following equations.
$\frac{C_{x+1}^{2x}}{C_{x-1}^{2x+1}}=\frac{2}{3}$ , $xϵN$

Answer 1

Given equation

$\frac{{}^{2x}{C}_{x+1}}{{}^{2x+1}{C}_{x-1}}=\frac{2}{3}$

We know that ${}^n{C}_r={}^n{C}_{n-r}$

And ${}^n{C}_r=\frac{n}{r}{}^{n-1}{C}_{r-1}$

Hence given equation can be written as

$\frac{{}^{2x}{C}_{2x-x-1}}{\frac{2x+1}{x-1}{}^{2x}{C}_{x-2}}=\frac{2}{3}$

$\Rightarrow \left(\frac{x-1}{2x+1}\right)\left(\frac{{}^{2x}{C}_{x-1}}{{}^{2x}{C}_{x-2}}\right)=\frac{2}{3}$

We know that $\frac{{}^n{C}_r}{{}^n{C}_{r-1}}=\frac{n-r+1}{r}$

$\Rightarrow \left(\frac{x-1}{2x+1}\right)\left(\frac{2x-x+1+1}{x-1}\right)=\frac{2}{3}$

$\Rightarrow \left(\frac{x+2}{2x+1}\right)=\frac{2}{3}$

$\Rightarrow 3(x+2)=2(2x+1)$

$\Rightarrow 3x+6=4x+2$

$\Rightarrow x=4$