Question

please solve this problem.....​

Answer 1

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.