Discrete Math Show that 3n n whenever n is an integer with

Discrete Math

Show that 3n < n! whenever n is an integer with n > 7.

Solution

Proof by Induction

n=8
3*8 = 24
8! = 40320
So, clearly 3*8 < 8! and the statement is true for base case

Let\'s assume that it holds good for n=k
3*k < k!

Now, we need to prove that for k+1 it holds good.
3*(k+1) = 3k + 3
Need to show that 3k + 3 < (k+1)!
(k+1)! = (k+1)(k!)
Let us say that k! = x
(k+1)! = (k+1)x = kx + x
-> 3k is less than x which is k! from Assumption Case
-> 3 is less than kx (Since k>7, multiplies any x=k! is greater than 3)
-> So, 3k + 3 together less than kx+x which is (k+1)!
3*(k+1) < (k+1)!

Hence, by Induction  3n < n! whenever n is an integer with n > 7.