Prove that $8^n > (2n-1)^2$ for $n>2$ by mathematical induction

Question

My attempt for this question:

Let n=3,

$8^3 > [2(3)-1]^2$

512 > 25 (True)

How do I show the steps for the n=k and n=k+1 for proving? I am not sure how to continue from here.

Answer 1

In order to prove this mathematical relation by induction for $n > 2$ we have to do two main steps. ( I'll assume $n \in N$)
First, let's prove that for $n=3$ the inequality is true. I'll save you the calculations as you've already done that.
Second, prove that, assuming the inequality is true for a certain $n=k$, it is automatically true for $n=k+1$.

( In this way, you know that if the inequality is true for $n=3$, it is automatically true for the next natural number $n=4$, but wait... You've just proved that the inequality is true for $n=4$, so you can apply the same reasoning and state that it is true for the next natural number $n=5$... et cetera.)

Thus, we know for sure that for $n=k\quad$ $8^k>(2k-1)^2$ and we have to demonstrate that $8^{k+1}>(2(k+1)-1)^2$. In order to tackle this demonstration, you have to use the inequality for $k$ that you hold for true. The approach will be , therefore, to reconduct the inequality for $k+1$ to a form similar to the inequality for $k$. After some algebraic passages we obtain $8^{k}>\frac{(2k+1)^2}{8}$. So you have to demonstrate that $8^k$ is strictly bigger than $\frac{(2k+1)^2}{8}$, but you already know for sure that $8^k$ is stricly bigger than $(2k-1)^2$. So if you demonstrate that $(2k-1)^2\ge\frac{(2k+1)^2}{8}$, you've proved your thesis.
(Note that if $\frac{(2k+1)^2}{8}$ is bigger than $\frac{(2k+1)^2}{8}$, $8^k$ could still be bigger than $\frac{(2k+1)^2}{8}$, so you would still have to demonstrate the thesis in another way.)
If you solve $(2k-1)^2\ge\frac{(2k+1)^2}{8}$ you obtain that it is true for all k.
So, here you are, you've demonstrated the inequality by induction for all $n>3$.