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(2n+1) (2n+3) (2n+5)......(4n-1) is equal to:
Option: 1
$\frac{(4 n) !}{2^n \cdot(2 n) !(2 n) !}$

Option: 2
$\frac{(4 n) ! n !}{2^n \cdot(2 n) !(2 n) !}$

Option: 3
$\frac{(4 n) ! n !}{(2 n) !(2 n) !}$

Option: 4
$\frac{(4 n) ! n !}{2^{n} !(2 n) !}$

Answers (1)

best_answer

E = (2n+1) (2n+3) (2n+5)......(4n-1)

Multiply numerator and denominator by (2n+2) (2n+4) .....(4n) & also by (2n)!

$\begin{aligned} E & =\frac{(2 n) !(2 n+1)(2 n+2)(2 n+3) \ldots \ldots(4 n-1) \cdot 4 n}{(2 n) !(2 n+2)(2 n+4) \ldots \ldots . .(2 n+2 n)} \\ & =\frac{(4 n) ! \times(n) !}{(2 n) ! 2^n[(n+1)(n+2) \ldots \ldots(2 n)] n !}=\frac{(n !) \cdot(4 n) !}{2^n \cdot((2 n) !)^2} \end{aligned}$

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