Originally Posted by

**pikminman** Hi I have an induction question. I think I have the thing done properly, I'm just not 100% sure.

The question is: prove that (3n)! <= 27^n (n!)^3

for all positive integers n.

This is what I have so far:

show for 1:

6<27

assume true for k where k is some positive integer

(3k)! <= 27^k (k!)^3

prove true for k+1

(3(k+1))! <= 27^(k+1) (k+1!)^3

(3k + 3)! <= (27^k)(27)((k!)^3)((k+1)^3)

((3k)!)(3k+1)(3k+2)(3k+3) <= (27^k)(27)((k!)^3)((k+1)^3)

THIS IS THE STEP I'M UNSURE ABOUT!!

I take out the original equation.. is this ok?

(3k+1)(3k+2)(3k+3) <= (27)(k+1)^3

27k^3 + 54k^2 + 33k + 6 <= 27k^3 + 81m^2 + 81m + 27

QED?