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 \(n \in \mathbb{N}*\).
