Question

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

• A. $n=2$
• B. $n=3$
• C. $n=4$
• D. $n=5$

Answer 1

The correct option is C $n=4$

Given ${}^{2n+1}{P}_{n-1}{:}^{2n-1}{P}_n=3:5$ which can be further simplified as
$\frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n-1)!}{(n-1)!}}=\frac{3}{5}$
$\Rightarrow \frac{(2n+1)\left(2n\right)}{(n+2)(n+1)\left(n\right)}=\frac{3}{5}$
$\Rightarrow$ $3n^2-11n-4=0$ .
On solving this quadratic equation we get roots as $4$ and $-\frac{1}{3}$.
Since $n$ is an integer,$n=4$.