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How would you go about proving something like $$2^n > 5n^2$$ for all integers n>8

The part I do not understand is how to solve for $n=k+1$

Also solving for $n=k+1$ here: $$2^n > 6n$$ for all integers n such that $n ≥ 5$

Any help is appreciated.

Important Edit Info: This question has been edited to fix a typo, so please ignore all answers stating that the problem is wrong, as they are referring to a previous, unedited version of the problem

Pellenthor
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TheReal7
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1 Answers1

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The inductive step could look something like this:

Say that for some k > 8 we already have that:

$$2^k > 5k^2$$

Then

$$2^{k+1} = 2(2^k) > 2 (5k^2) = 10k^2 = 5k^2 + 5k^2 \geq 5k^2 + 5(2k+1) \geq 5k^2 + 10k + 5 = 5(k+1)^2$$

where we used the fact that $k^2 \geq 2k + 1$ for all $k \geq 3$ (Why? - to see this let $k \geq 3$ then: $k^2 = k*k \geq 3k = 2k + k \geq 2k +1$)

We conclude that:

$$2^{k+1} > 5(k+1)^2$$

Pellenthor
  • 955