please solve this problem.....
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Step-by-step explanation:
(2n+1)!/(2n+1-(2n-1))!
:
(2n-1)!/(2n-1-n)!=3/5
=>(2n+1)!/2! : (2n-1)!/(n-1)!=3/5
=>(2n+1)!/2 : (2n-1)!(2n)(2n+1)/(n-1)!(2n)(2n+1)=3/5
=>(2n+1)!/2 : (2n+1)!/(n-1)!(2n)(2n+1)=3/5
=>(n-1)!(2n)(2n+1)=6/5
=>n!2(2n+1)=6/5
=>n!(2n+1)=3/5
=>2ⁿ n!(2n+1)/2ⁿ=3/5
=>(2n+1)!/2ⁿ=3/5
which is not possible as 2ⁿ should be a term in the form 2ⁿ,but here it is 5.
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