Can you help me?
Prove that 3^n > n^3 for all positive integers n.
I first tried n= 1, which showed 3>1
n= 2, 9>8
but n = 3 became 27 = 27, which is not true, but everything else works afterwards.
I was given the hint to use induction. So I tried this:
3^(n+1) > (n+1)^3
3*(3^n) > n^3 + 3n^2 + 3n + 1
What do I do next?
Suppose for some which was shown.
For the inductive step, you want to show that .
Since it is known that then .
Also, leads to , so and it follows that . Rearranging, it is evident that .
So applying the above revelations,
It is hypothesized that , so
Thus, , and by induction, is true for all .
Thank you for your help. But I need a little more explantion.
I don't quite know where or why you did this particular line: k^3 >= 4k^2>3k^2
Then why did you do the two lines after the above line?
I'll try to be clearer:
I wanted to make it clear that and .
Using both of these facts, then , so which is the expansion of .
So all of the above sets up:
and since , then . in other words, , which is what we wre trying to show.