Question

please solve this with explanation​

Answer 1

Step-by-step explanation:

LHS

(2n!/n!)     ------- (1)

1st solve for 2n!.

2n!

=> 1 * 2 * 3 * 5 * 6 * .... * n * (n+1) * .... * (2n-1) * (2n)

=> [1 * 3 * 5 * ... * (2n-1)] * [2 * 4 * 6 * ... * (2n)]

=> [1 * 3 * 5 * ... * (2n-1)] * [ (2*1) * (2*2) * (2*3) * (2*4) * ... * (2*n)]

=> [1 * 3 * 5 * ... * (2n-1)] * [ (2 * 2 * 2 ...) * (1 * 2 * 3 * 4 * 5 * .... * n)]

=> [1 * 3 * 5 * ... * (2n-1)] * [2^n * n!]

Place the value of 2n! in (1), we get

=> [1 * 3 * 5 * .... (2n - 1) * [2^n * n!]/n!

=> [1 * 3 * 5 * ... (2n - 1) * 2^n

RHS

#Hope my answer helped you!