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