• Support PF! Buy your school textbooks, materials and every day products Here!

Induction proof verification ##2^{n+2} < (n+1)!## for all n ##\geq 6##

  • Thread starter ciencero
  • Start date
1
0
$2^{n+2} < (n+1)!$ for all n $\geq 6$

Step 1: For n = 6,

$256 < 5040$.

We assume

$2^{k+2} < (k+1)!$

Induction step:

$2 * 2^{k+2} < 2*(k+1)!$

By noting $2*(k+1)! < (k+2)!$

Then $2^{k+3} < (k+2)!$
 

Answers and Replies

33,564
9,295
Looks fine.
n=6 is not the first place where the inequality is true, by the way.
 
32,928
4,637
$2^{n+2} < (n+1)!$ for all n $\geq 6$

Step 1: For n = 6,

$256 < 5040$.

We assume

$2^{k+2} < (k+1)!$

Induction step:

$2 * 2^{k+2} < 2*(k+1)!$

By noting $2*(k+1)! < (k+2)!$

Then $2^{k+3} < (k+2)!$
@ciencero, at this site, use double $ characters at each end for standalone LaTeX, or double # characters at each end for inline LaTeX.
 
WWGD
Science Advisor
Gold Member
5,009
2,240
It seems clear the RH side will eventually dominate. LH is being multiplied by 2 from nth to (n+1)st term while RH side is being multiplied by increasingly larger factors.
 

Related Threads for: Induction proof verification ##2^{n+2} < (n+1)!## for all n ##\geq 6##

Replies
15
Views
75K
Replies
12
Views
2K
  • Last Post
Replies
9
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
8
Views
942
  • Last Post
Replies
5
Views
3K
Replies
4
Views
5K
Replies
5
Views
850
Top