Stuck on proof by induction

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October 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.

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

Basis for Induction

$3*2^5 < 5!$

$96 < 120$

Induction Step

We assume $3*2^n <n!$ 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!
October 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.

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

Basis for Induction

$3*2^5 < 5!$

$96 < 120$

Induction Step

We assume $3*2^n <n!$ 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
October 17th 2010, 08:21 AM
Plato

Recall that $n>1$ then $n+1>2$ .

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

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