# Prove by induction

1. Sep 15, 2009

### rbnphlp

3n>n3 where n >3

I know I have to use proof by induction to solve this.
assume for f(4)is true
for f(n+1)=3n+1>3n3

However after that I dont have a clue of how to getting it into (n+1)3
Any hints will be greatly apperciated

2. Sep 15, 2009

Let

$$A_n$$

be the statement

$$3^n > n^3$$

You want to show $$A_n$$ is true for $$n \ge 4$$.

It shouldn't be hard to show $$A_4$$ is true. Assume it is true for $$k \ge 4$$.

Now

$$3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3$$

Now play with the terms on the right, making the new expressions ever smaller, to build up to the expansion of $$(k+1)^3$$.

(Hint for a start: You know by hypothesis $$k \ge 4 > 3$$, so
$$k^3 = k \cdot k^2 > 3k^2$$
which is the second term in the expansion of $$(k+1)^3$$)

Last edited by a moderator: Sep 15, 2009
3. Sep 16, 2009

### rbnphlp

This is th best I could get at:

3n>n3
For n+1=
3n+1>3n

given n>3 , I found the following:

• n3>3n2
• n2>3n
• n-2>1
with the best intentions I did the following:

Added all them up and got
n3>3n2+3n+1-n2-n+2--(1)
3n+1-2n3>n3--(2)

from 1 and 2
3n+1-2n3>3n2+3n+1-n2-n+2

3n+1+n2+n-2-n3>(n+1)3

But after that im stuck it would be great some one could help. thnx

4. Sep 16, 2009

### rbnphlp

anyone?

5. Sep 16, 2009

### HallsofIvy

Staff Emeritus
Statdad showed that you could get to $k^3+ k^3+ k^3$ and you want to compare that to $(k+1)^3= k^3+ 3k^2+ 3k+ 1$. Can you show that $k^3> 3k^2$ when k> 3? Can you show that $k^3> 3k+ 1$ when k> 3?

6. Sep 17, 2009

### rbnphlp

nope I cant .