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 .