Proving n3 + n < 3n for n >= 4 using Mathematical Induction

[lovemake1]

Homework Statement

show n³ + n < 3ⁿ for all n >= 4

Homework Equations

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?

 

[szynkasz]

Yes it does.

 

[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 I am 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.

 

[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?

 

[HallsofIvy]

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

 

[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?

 

[szynkasz]

Yes, you are.