Prove, by mathematical induction, that n!>3^n for nEN and n>=7?

Answer 1

Base case: When n=7, n! = 5040 and 3^n = 2187.
So n!>3^n for our base case.

Induction step: Given that n!>3^n,
need to show that (n+1)!>3^(n+1).

We are given that n!>3^n
and n is a positive integer >= 7.

so 3*(n!) > 3*(3^n)
so 3*(n!) > 3^(n+1)
Since n >= 7, (n+1) must be greater than 3.
Hence, (n+1)*n! > 3*(n!)
so (n+1)! > 3^(n+1)
which is what we needed to prove.

Hope that helps!

Answer 2

Assume true for n = k.
P(k) is the proposition that:-
k! > 3^k for k ≥ 7
Have to prove true for k = 7 and k > 7.

Consider P(7)
LHS = 7! = 5040
RHS = 3^7= 2187
Thus P(7) is true

Consider P(k + 1)
(k + 1)! > 3^(k + 1)------(have to prove this)
Now k! > 3^k is true
(k + 1).k! > (k + 1).3^k
(k + 1)! > (k + 1).3^k
Now have to show that:-
(k + 1).3^k > 3^(k + 1)
(k + 1).3^k - 3^(k + 1) > 0
3^k.(k + 1 - 3) > 0
3^k.(k - 2) > 0
Now k ≥ 7 so this is true.
Thus P(k + 1) is true.

P(k) is true
P(1) is true
P(k + 1) is true
Thus P(n) is true.
(k + 1).3^k = 3^k.(k + 1)

Answer 3

Suppose n≥7 and n!>3^n. Then (n+1)! = (n+1)*n! > 3*n! > 3*3^n = 3^(n+1), so (n+1)! > 3^(n+1). Since 7! = 5040 > 2187 = 3^7, this inequality holds for n=7, and we already showed that n!>3^n implies (n+1)!>3^(n+1), so by induction, it holds for all n≥7. Q.E.D.