# Stuck on proof by induction

• Oct 17th 2010, 09:13 AM
dtwazere
Stuck on proof by induction
Hey, Stuck on this proof by induction question. I can get so far but not sure where to go after.

$3*2^n < n!$ when $n>4$

Basis for Induction

$3*2^5 < 5!$

$96 < 120$

Induction Step

We assume $3*2^n when $n>4$

We need to prove that $3*2^n^+^1 < (n+1)!$

$(n+1)! = (n+1)n!$

$(3*2^n)(n+1)<(n+1)n!$

$(3*2^n)(n+1)<(n+1)!$

This is as far as I can get :( Any help would be appreciated!
• Oct 17th 2010, 09:21 AM
tonio
Quote:

Originally Posted by dtwazere
Hey, Stuck on this proof by induction question. I can get so far but not sure where to go after.

$3*2^n < n!$ when $n>4$

Basis for Induction

$3*2^5 < 5!$

$96 < 120$

Induction Step

We assume $3*2^n when $n>4$

We need to prove that $3*2^n^+^1 < (n+1)!$

$(n+1)! = (n+1)n!$

$(3*2^n)(n+1)<(n+1)n!$

$(3*2^n)(n+1)<(n+1)!$

This is as far as I can get :( Any help would be appreciated!

$3\cdot 2^{n+1}=2(3\cdot 2^n)<2\cdot n! < (n+1)n!=(n+1)!$ , the last inequality being true since $n+1>2$ ...

Tonio
• Oct 17th 2010, 09:21 AM
Plato
Recall that $n>1$ then $n+1>2$.

Start with $3\cdot 2^{n+1}=[3\cdot 2^{n}](2)<[n!](?)$