If ${ }^{2\mathrm{n}+1}{\mathrm{P}}_{\mathrm{n}-1}{:}^{2\mathrm{n}-1}{\mathrm{P}}_{\mathrm{n}}=3:5$ then the value of n is equal to

Solution:

We have, ${ }^{2\mathrm{n}+1}{\mathrm{P}}_{\mathrm{n}-1}{:}^{2\mathrm{n}-1}{\mathrm{P}}_{\mathrm{n}}=3:5$

$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{{ }^{2\mathrm{n}+1}{\mathrm{P}}_{\mathrm{n}-1}}{2\mathrm{n}-1_{P_n}}=\frac{3}{5}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{(2\mathrm{n}+1)!}{(\mathrm{n}+2)!}\times \frac{(\mathrm{n}-1)!}{(2\mathrm{n}-1)!}=\frac{3}{5}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{(2\mathrm{n}+1)\left(2\mathrm{n}\right)}{(\mathrm{n}+2)(\mathrm{n}+1)\mathrm{n}}=\frac{3}{5}\\ \\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}10(2\mathrm{n}+1)=3(\mathrm{n}+2)(\mathrm{n}+1)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3{\mathrm{n}}^2-11\mathrm{n}-4=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(\mathrm{n}-4)(3\mathrm{n}+1)=0\Rightarrow \mathrm{n}=4\end{array}$