Question

Using Mathematical Induction,

Prove that : $\mathsf{2^n \: > \: n^3, \: where \: n > 9}$

Answer 1

Here it's given that n > 9. So on considering P(1) doesn't get the result for the proof. So we first consider P(10).

We have to prove,

$P(n):\ 2^n>n^3,\ \ n\in\mathbb{N},\ \ n>9$

So, consider P(10).

$2^{10}=1024>1000=10^3\ \ \ \ \ \Longrightarrow\ \ \ \ \ 2^{10}>10^3$

So P(10) holds true. Thus we can assume P(k) is true.

$\text{Let}\ \ P(k):\ 2^k>k^3,\ \ k\in\mathbb{N},\ \ k>9$

So, let  $2^k=k^3+x,\ \ \text{where \ x\in\mathbb{Z^+} .}$

This is taken as equation no. 1.

Now, consider P(k + 1).

From equation no. 1, taking LHS,

$2^{k+1}=2^k\cdot 2=(k^3+x)2=2k^3+2x$

Taking RHS,

$(k+1)^3=k^3+3k^2+3k+1$

Now we just have an assumption that both LHS and RHS possess the considered inequality. Then we have many operations with it until getting another equation which always holds true.

So, assume,

$\begin{aligned}&2k^3+2x>k^3+3k^2+3k+1\\ \\ \Longrightarrow\ \ &k^3+k^3+2x>k^3+3k^2+3k+1\\ \\ \Longrightarrow\ \ &k^3+2x>3k^2+3k+1\end{aligned}$

But before the continuation of this, we just take,

$k^3>3k^2+3k+1\ \ \ \ \ \Longrightarrow\ \ \ \ \ k^3>3k(k+1)+1$

But it's true ∀k ≥ 4. How?!

As did earlier, we just assume this one holds the inequality, so,

$\begin{aligned}&k^3>3k(k+1)+1\\ \\ \Longrightarrow\ \ &k^3-1>3k(k+1)\\ \\ \Longrightarrow\ \ &(k-1)(k^2+k+1)>3k(k+1)\\ \\ \Longrightarrow\ \ &(k-1)(k(k+1)+1)>3k(k+1)\\ \\ \Longrightarrow\ \ &k(k-1)(k+1)+k-1>3k(k+1)\\ \\ \Longrightarrow\ \ &k(k-1)(k+1)-3k(k+1)+k-1>0\\ \\ \Longrightarrow\ \ &k(k-4)(k+1)+k-1>0\end{aligned}$

The final step always holds true ∀k ≥ 4.

So, come to our inequality.

$k^3+2x>3k^2+3k+1$

This inequality is exactly true since all of the following are considered or true,

$k\geq 9>4,\ \ x\in\mathbb{N},\ \ k^3>3k^2+3k+1$

Thus we get that  P(k + 1)  is also true.

Hence Proved!