If `^2n+1P_(n-1):^(2n-1)P_n=3:5,fin dndot`

Explanation

We have
`""^(2n+1)P_(n-1):""^(2n-1)P_(n)=3:5`
`rArr ((2n+1)!)/({(2n+1)-(n-1)}!):((2n-1)!)/({(2n-1)-n}!)=(3)/(5)`
`rArr ((2n+1)!)/((n+2)!):((2n-1)!)/((n-1)!)=(3)/(5)`
`rArr ((2n+1)!)/((n+2)!)xx((n-1)!)/((2n-1)!)=(3)/(5)`
`rArr ((2n+1)xx2nxx[(2n-1)!])/((n+2)xx(n+1)xxnxx[(n-1)!])xx((n-1)!)/((2n-1)!)=(3)/(5)`
`rArr 10(2n+1)=3(n+2)(n+1)`
`rArr 20n+10=3(n^(2)+3n+2)`
`rArr 3n^(2)-11n-4=0`
`rArr (n-4)(3n+1)=0 rArr n =4 [because n ne (-1)/(3), " as n cannot be negative"].` ltbegt Hence, `n=4.`