Question

Solve the following inequalities.
$C_4^{x-1}-C_3^{x-1}-\frac{5}{4}{C}_2^{x-2}<0,xϵN$

Answer 1

$C_4^{x-1}-C_3^{x-1}-\frac{5}{4}{C}_2^{x-2}<0,xϵN$

$\Rightarrow C_4^{x-1}+C_3^{x-1}<\frac{5}{4}{C}_2^{x-2}$

$\Rightarrow C_4^{x-1}+1+\left(C_3^{x-1}\right).<\frac{5}{4}{C}_2^{x-2}$

${\Rightarrow }^{x+1-1}{C}_4.<\frac{5}{4}{C}_2^{x-2}$ by formula $C_r^n+C_{r-1}^n{=}^{n+1}{C}_r$

$\Rightarrow C_4^x<\frac{5}{4}{C}_2^{x-2}$

$\Rightarrow \frac{x!}{\left(4\right)!(x-4)!}<\frac{5}{4}\frac{(x-2)!}{\left(2\right)!(x-2-2)!}$

$\Rightarrow \frac{x!}{(4!)(x-4)!}<\frac{5}{4}\frac{(x-2)!}{(x-4)!\times 2!}$

$\Rightarrow \frac{x(x-1)(x-1)(x-2)!}{4!}<\frac{5}{4}\frac{(x-2)!}{2}$

$\Rightarrow \frac{x(x-1)\times 2\times 4}{4!5}<1$

$\Rightarrow \frac{x(x-1)\times 2\times 4}{4\times 3\times 2\times 1\times 5}<1$

$\Rightarrow x^2-x<15$

$\Rightarrow x^2-x-15<0$

Let $x^2-x-15=0$ Here

So $(x-\frac{(1-\sqrt{61})}{2})(x-\frac{(1-\sqrt{61})}{2})$

Hence

$xϵ(\frac{1-\sqrt{61}}{2},\frac{1+\sqrt{61}}{2})$

$x=\frac{+1\pm \sqrt{1-4\times (-15)}}{2}$

$\frac{+1\pm \sqrt{1+60}}{2}=\frac{+1\pm \sqrt{61}}{2}$