Question

Solve the following inequalities.
$5C_3^n<C_4^{n+2},nϵN$

Answer 1

Solve $5C_3^n<C_4^{n+2},nϵN$

Sol${}^n$

$5C_3^n<C_4^{n+2}$

$\Rightarrow \frac{5\left(n\right)!3!}{(n-3)!}<\frac{(n+2)!}{(n+2-4)!(4!)}$

$\Rightarrow \frac{5(n!)}{(n-3)!3!}<\frac{(n+2)(n+1)\left(n\right)!}{(n-2)!\left)\right(4!)}$

$\Rightarrow \frac{5}{3!(n-3)!}<\frac{(n+2)(n+2)}{(n-2)(n-3)!}4!$ as $4!=4\times 3!$

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

$\Rightarrow 20(n-2)<(n+2)(n+1)$

$\Rightarrow (n+2)(n+1)>20(n-2)$

$\Rightarrow n^2+3n+2-20(n-2)>0$

$\to n^2+3n+2-20n+40>0$

$\Rightarrow n^2-17n+42>\mathrm{0...}\left(i\right)$

Let $n^2-17n+42=0$

then $n=\frac{+17\pm \sqrt{289-4\times 1\times 42}}{2}$

$n=\frac{+17\pm \sqrt{121}}{2}=\frac{17\pm 11}{2}$

then $n=\frac{17+11}{2}$ and $n=\frac{17-11}{2}$

$n=14$ and $n=3$

So, eq (I) because $(n-3)(n-14)>0$

then $nϵ(-\infty ,3)\cup (14,\infty ).$