If n is any positive integer, then 1 2 n ( P n 2 n ) =

Question

If $n$ is any positive integer, then $\frac{1}{2^{n}} ({}^{2 n}P_{n}) =$

Smart Achievers · Accepted Answer

(3)

$\frac{1}{2^{n}} ({}^{2 n}P_{n}) = \frac{1}{2^{n}} (\frac{(2 n)!}{n!})$

$= \frac{(2 n) (2 n - 1) (2 n - 2) \dots 3 \cdot 2 \cdot 1}{2^{n} \times n!}$

$= \frac{[(2 n) (2 n - 2) \dots 6 \cdot 4 \cdot 2] [(2 n - 1) (2 n - 3) \dots 5 \cdot 3 \cdot 1]}{2^{n} \times n!}$

$= \frac{2^{n} (n (n - 1) \dots 3 \cdot 2 \cdot 1) (1 \cdot 3 \cdot 5 \dots (2 n - 1))}{2^{n} \cdot (n!)}$

$= \frac{2^{n} (n!) (1 \cdot 3 \cdot 5 \dots (2 n - 1))}{2^{n} \cdot (n!)}$

$= 1 \cdot 3 \cdot 5$ ...... $(2 n - 1)$

Want Us to Email you About Special Offers & Updates?