## Solving Lesson 6 Page 29 Math Learning Topic 10 – Kite>

Topic

Prove $${n^n} > {(n + 1)^{n – 1}}$$ for all $$n \in \mathbb{N}*,n \ge 2.$$

Solution method – See details

Inductive method: Prove that the statement is true with $$n \ge p$$

Step 1: Check the statement is true for $$n = p$$

Step 2: Suppose the proposition is true for natural numbers $$n = k \ge p$$ and prove the statement true for $$n = k + 1.$$ Conclusion.

Detailed explanation

Step 1: When $$n = 2$$ we have $${2^2} > {(2 + 1)^{2 – 1}}$$ or $$4 > 3$$ obviously true

So the inequality holds for $$n = 2$$

Step 2: With k being an arbitrary positive integer that the inequality is true, we must prove the inequality is true for k+1, that is:

$${(k + 1)^{k + 1}} > {(k + 1 + 1)^{k + 1 – 1}}$$ or $${(k + 1)^{k + 1} } > {(k + 2)^k}$$

Indeed, by the assumption of induction we have:

$${k^k} > {(k + 1)^{k – 1}}$$

I guess

$${k^k}{(k + 1)^{k + 1}} > {(k + 1)^{k – 1}}{(k + 1)^{k + 1}} = {( k + 1)^{k – 1 + k + 1}} = {(k + 1)^{2k}}$$

Where $${(k + 1)^{2k}} = {\left[ {{{(k + 1)}^2}} \right]^k} = {\left( {{k^2} + 2k + 1} \right)^k} > {\left( {{k^2} + 2k} \right)^k}$$

$$\Rightarrow {k^k}{(k + 1)^{k + 1}} > {\left( {{k^2} + 2k} \right)^k} = {\left[ {k.(k + 2)} \right]^k} = {k^k}. {(k + 2)^k}$$

$$\Rightarrow {(k + 1)^{k + 1}} > {(k + 2)^k}$$

So the inequality holds for k+1. Thus, by the principle of mathematical induction, the inequality holds for all $$n \in \mathbb{N}*$$.