Proof by Induction: 3^n >= 1+2^n

[eku_girl83]

Prove that for all n in the natural numbers 3^n greater than or equal to 1+2^n.

Here's my start:
3^1 greater than or equal to 1+2^1, so the statement is true for n=1.
Assume that for some n, 3^n greater than or equal to 1+2^n
Then 3^n+3^n+3^n greater than or equal to 1+2^n +3^n+3^n.
It follows that 3^(n+1) greater than or equal to 1+2^n+2*3^n.


Where do I go from here? I still need to show that 1+2^n+2*3^n is greater than or equal to 1+2^(n+1)

Thanks!

 

[NateTG]

You seem to be having a little trouble formatting. There's a thread on using latex here:
https://www.physicsforums.com/showthread.php?t=8997&highlight=Latex

You know, from the inductive hypothesis that:
3^n \geq 1+2^n
perhaps you could use that?

 

[mathman]

To show 3ⁿ⁺¹>1+2ⁿ⁺¹
subtract the n expressions from both sides, and you will have to show
2*3ⁿ>2ⁿ
That looks obviously true,since 3>2.

 

[somy]

good work mathman!

 

[Fredrik]

So you'd like to prove that

3^n\geq 1+2^n

using induction. The statement is obviously true for n=1, so all you have to do is to prove that if it's true for n=k, it's also true for n=k+1 (whatever k is).

So you should start like this:

3^{k+1}=3\cdot 3^k\geq 3(1+2^k)\geq\dots

Now all you have to do is to show that this is

\geq 1+2^{k+1}

This is very easy. (Hint: 3>2>1).