Given,

$\frac{{}^{2n+1}P_{n-1}}{{}^{2n-1}P_n}=\frac{11}{21}$

$\Rightarrow \frac{{\displaystyle \frac{\left(2n+1\right)!}{\left(n+2\right)!}}}{{\displaystyle \frac{\left(2n-1\right)!}{\left(n-1\right)!}}}=\frac{11}{21}$

$\Rightarrow 21\times \frac{\left(2n+1\right)2n\left(2n-1\right)!}{\left(n+2\right)\left(n+1\right)n\left(n-1\right)!}=11\times \frac{\left(2n-1\right)!}{\left(n-1\right)!}$

$\Rightarrow 21\times 2\left(2n+1\right)=11\left(n+2\right)\left(n+1\right)$

$\Rightarrow \frac{\left(2n+1\right)}{\left(n+2\right)\left(n+1\right)}=\frac{11}{42}$

Now on comparing both side we get, $n=5$

So, the value of $n^2+n+15=5^2+5+15=45$