# Mathematical Induction

1. Sep 29, 2012

### lovemake1

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

show n3 + n < 3n for all n >= 4

2. Relevant equations

3. The attempt at a solution

I.H : n3 + n < n for all n >= 4

3(n3 + n) < 3(3n)
then (3n+1) = 3 x 3 n
> 3((n3) + n ) by I.H
> (n+1)3 + (n+1)

if we show 3(n3 + n ) - [(n+1)3 + (n+1)] > 0 by subbin in 4 which is n >= 4, does it suffice as proof?

Last edited: Sep 29, 2012
2. Sep 29, 2012

### szynkasz

Yes it does.

3. Sep 29, 2012

### lovemake1

> 3((n3) + n ) by I.H
I am still unsure if I got it right....

I said in my induction hypothesis that, n3 + n < 3n
but while im trying to prove that p(k+1) works for all k,
I think I am using assuming this line.... which is the p(k+1) that I am trying to prove. 3(n3 + n) < 3(3n) by Induction Hypothesis.

so technically I am using what I need to prove to prove my question... ahaha
im just confused, can someone check my work please, thank you.

4. Sep 29, 2012

### szynkasz

$$3^{n+1}=3\cdot 3^n>3(n^3+n)$$
so you need to prove:
$$3(n^3+n)>(n+1)^3+(n+1)$$
and then:
$$3^{n+1}>(n+1)^3+(n+1)$$
Is this what you mean?

5. Sep 29, 2012

### HallsofIvy

Staff Emeritus
I don't see where you have proven it for n= 6.

6. Sep 29, 2012

### lovemake1

ok I think i figured out the trick.
I expand both sides, but on the left side i make
n3 + n3 + n3 + 3n and compare it to the right side term by term to prove that its greater for all n > 4.

am I right with this approach?

7. Sep 30, 2012

### szynkasz

Yes, you are.