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

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

$\frac{(2n+1)\left(2n\right)(2n-1)!}{(n+2)!}\times \frac{(n-1)!}{(2n-1)!}=\frac{3}{5}$

$\frac{(2n+1)\left(2n\right)}{(n+2)(n+1)\left(n\right)(n-1)!}\times \frac{(n-1)!}{1}=\frac{3}{5}$

$\frac{2(2n+1)}{(n+2)(n+1)}=\frac{3}{5}\Rightarrow 10(2n+1)=3(n^2+3n+2)$

$\begin{array}{l}\Rightarrow 3n^2+9n+6-20n-10=0\\ \Rightarrow 3n^2-11n-4=0\\ \Rightarrow 3n^2-12n+n-4=0\\ 3n(n-4)+1(n-4)=0\\ n=4,n=\frac{-1}{3}\end{array}$