# Stuck on proof by induction

• Oct 17th 2010, 08:13 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.

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

Basis for Induction

$\displaystyle 3*2^5 < 5!$

$\displaystyle 96 < 120$

Induction Step

We assume $\displaystyle 3*2^n <n!$ when $\displaystyle n>4$

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

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

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

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

This is as far as I can get :( Any help would be appreciated!
• Oct 17th 2010, 08:21 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.

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

Basis for Induction

$\displaystyle 3*2^5 < 5!$

$\displaystyle 96 < 120$

Induction Step

We assume $\displaystyle 3*2^n <n!$ when $\displaystyle n>4$

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

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

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

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

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

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

Tonio
• Oct 17th 2010, 08:21 AM
Plato
Recall that $\displaystyle n>1$ then $\displaystyle n+1>2$.

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