# I need help with this induction proof

1. Dec 11, 2012

### snapgilmore

1. The problem statement, all variables and given/known data

Prove that 2^n > n^3 for every integer n >= 10

2. Relevant equations

a) Prove for 10,
b) assume for k
c) inductive step for k+1

3. The attempt at a solution

a)
2^10 = 1024
10^3 = 1000
1024>1000

b)Assume
2^k > k^3

c) 2^(k+1) > (k+1)^3
(2)(2^k) > (k+1)^3
(2^k) > ((k+1)^3)/2

and from here I'm having trouble solving for k >= 10.
Any help would be appreciated, thanks.

2. Dec 12, 2012

### Staff: Mentor

The inequality above is what you need to show. You can't just assert it.
Your work should look like this:
2k+1 = 2 * 2k > 2 * k3 ...

You then need to show that 2k3 > (k + 1)3, at least for k ≥ 10.

It suffices to show that 2k3/(k + 1)3 > 1, which can be done using calculus.

Last edited: Dec 12, 2012
3. Dec 12, 2012

### snapgilmore

Can you elaborate on how its enough to show (2k^3)/(k+1)^3 >1 ? I thought you needed to end up at k >=10?

4. Dec 12, 2012

### SammyS

Staff Emeritus
You showed that the inequality holds for n = 10.

Then you show that if the inequality holds for n = k, where k is any number greater than or equal to 10, then the inequality holds for n = k+1 .

By doing this, you show that if it's true for n=10, then it's true for n=11 .
If it's true for n=11, then it's true for n=12 .

If it's true for n=12, then it's true for n=13 .

If it's true for n=13, then it's true for n=14 .

If it's true for n=14, then it's true for n=15 .

...​
So assume that 2k > k3, for some integer k ≥ 10 .

From that show that 2k+1 > (k+1)3 .

5. Dec 12, 2012

### Staff: Mentor

(2k3)/(k+1)3 >1 is equivalent to 2k3 > (k + 1)3.