Question

show that (2n)!/n!=2^n[1.3.5...(2n-1)]

Answer 1

$LHS=\frac{(2n)!}{n!}$

we know, x! = x(x - 1)(x - 2)(x - 3)(x - 4)......1

similarly, n! = n(n - 1)(n - 2)(n - 3).......1 ...(1)

(2n)! = 2n(2n - 1)(2n - 2)(2n - 3)(2n - 4).......1

= {2n(2n - 2)(2n - 4)(2n - 6)(2n - 8).....2}{(2n - 1)(2n - 3)(2n - 5).....3.1}

=[2^n{n(n - 1)(n - 2)(n - 3)(n - 4)......1 ]{1.3.5 .....(2n -5)(2n - 3)(2n - 1)}

=( 2^n n!) {1.3.5......(2n - 1)} [ from equation (1)

now, $\frac{(2n)!}{n!}=\frac{( 2^n n!)\{1.3.5......(2n - 1)\}}{n!}$

= $2^n[1.3.5......(2n-1)$ = RHS