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



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