Question

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

Answer 1

We have,

${}^{2n+1}{P}_{n-1}{:}^{2n-1}{P}_n=3:8$

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

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

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

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

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

$\frac{(2n+1)2}{(n+2)(n+1)\times 3}=\frac{1}{5}$

$\frac{4n+2}{3(n^2+2n+n+2)}=\frac{1}{5}$

$20n+10=3(n^2+3n+2)$

$20n+10=3n^2+9n+6$

$3n^2+9n+6-20n-10=0$

$3n^2-11n-4=0$

$3n^2-(12-1)n-4=0$

$3n^2-12n+n-4=0$

$3n(n-4)+1(n-4)=0$

$(n-4)(3n+1)=0$

If $n-4=0$ then, $n=4$ $n\in N$

If $3n+1=0$ then, $n=-\frac{1}{3}$ $n\notin N$

Hence, $n=4$ is the answer.