Question

If ${}^{2n+1}{P}_{n-1}:{}^{2n-1}{P}_n=11:21$, then $n^2+n+15$ is equal to :

Answer 1

Answer: 45

$\frac{(2n+1)!(n-1)!}{(n+2)!(2n-1)!}=\frac{11}{21}$
$\Rightarrow \frac{(2n+1)(2n)}{(n+2)(n+1)n}=\frac{11}{21}$
$\Rightarrow \frac{2n+1}{(n+1)(n+2)}=\frac{11}{42}$
$\Rightarrow n=5$
$\Rightarrow n^2+n+15$
$=25+5+15=45$