Question

If ${}^{(}{}^2{}^n{}^{+}{}^1{}^{)}{P}_{{}_n}{}_{-}{}_1:{}^{(}{}^2{}^n{}^{-}{}^1{}^{)}{P}_{{}_n}=$ $3:5$ then n$=$

• A. $4$
• B. $5$
• C. $6$
• D. $3$

Answer 1

The correct option is A $4$

$\frac{{}^{2n+1}{P}_{n-1}}{{}^{2n-1}{P}_n}=\frac{3}{5}$
$\left(\frac{(2n+1)!}{(n+2)!}\right)/\left(\frac{(2n-1)!}{(n-1)!}\right)=\frac{3}{5}$
$\frac{(2n+1)!(n-1)!}{(n+2)!(2n-1)!}=\frac{3}{5}$
$\frac{(2n+1)\left(2n\right)}{(n+2)(n+1)\left(n\right)}=\frac{3}{5}$
$\frac{(2n+1)\left(2\right)}{(n+2)(n+1)}=\frac{3}{5}$
$\Rightarrow \frac{10}{3}=\frac{(n+1)(n+2)}{2n+1}$
$\Rightarrow 20n+10=(n^2+3n+2)\left(3\right)$

$\Rightarrow 3n^2+9n+6-20n-10=0$

$\Rightarrow 3n^2-11n-4=0$

$\Rightarrow (3n+1)(n-4)=0$

$\Rightarrow n=4,-1/3$

$\Rightarrow n=4$