Question

Prove that the expression (3n)!/(3!)n is an integer for all values of n greater than and equal to 0

Answer 1

Hey

Here 'n' only admits Natural Number Values !!

$= > \frac{(3n) !}{3 !n} = \frac{3! {n}^{3}.n!}{3 !n} = {n}^{2} n!$
Hence, we see the Resultant is obviously Natural for all permissible values of n ^^"