Question

If P(2n-1,n): P(2n+1,n-1)= 22:7 find n.

Answer 1

We have,
P(2n-1,n): P(2n+1, n-1)= 22:7
$\Rightarrow \frac{P(2n-1,n)}{P(2n+1,n-1)}=\frac{22}{7}\Rightarrow \frac{\frac{(2n-1)!}{(2n-1-n)!}}{\frac{(2n+1)!}{[2n+1-(n-1\left)\right]!}}=\frac{22}{7}\Rightarrow \frac{(2n-1)!\times (n+2)!}{(n-1)!(2n+1)!}=\frac{22}{7}\Rightarrow \frac{(2n-1)!\times (n+2)(n+1).n.(n-1)!}{(n-1)!(2n+1).2n.(2n-1)!}=\frac{22}{7}\Rightarrow \frac{n(n+1)(n+2)}{2n(2n+1)}=\frac{22}{7}\Rightarrow \frac{(n+2)(n+1)}{2(2n+1)}=\frac{22}{7}\Rightarrow \frac{n^2+n+2n+2}{4n+2}=\frac{22}{7}\Rightarrow 7(n^2+3n+2)=22\times (4n+2)\Rightarrow 7n^2+21n+14=88n+44\Rightarrow 7n^2+21n-88n+14-44=0\Rightarrow 7n(n-10)+3(n-10)=0\Rightarrow 7n^2-70n+3n-30=0\Rightarrow 7n(n-10)+3(n-10)=0\Rightarrow n=10[\because 7n+3\ne 0]$