Question 383270
 (2n+1)!/(n+2)! * (n-1)!/(2n-1)!=3/5 what is n?
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Rearrange so you can see whats happening:
[(2n+1)!/(2n-1)!]*[(n-1)!/(n+2)!] = 3/5
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Cancel factors that are common to a numerator and a denominator:
 [(2n+1)(2)/1]*[1/(n+2)(n+1)] = 3/5
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 [4n+2]/[n^2+3n+2] = 3/5
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 5(4n+2) = 3(n^2+3n+2)
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20n+10 = 3n^2+9n+6
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3n^2-11n-4 = 0
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3n^2-12n+n-4 = 0
Factor:
3n(n-4)+(n-4) = 0
(n-4)(3n+1) = 0
Positive solution:
n = 4
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Cheers,
Stan H.