Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.4 - Mathematical Induction - Exercise Set - Page 1086: 41


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Work Step by Step

Let ${{S}_{n}}$ $:{{n}^{2}}>2n+1$ Suppose ${{S}_{n}}$ be a statement involving the positive integer n. If 1. ${{S}_{3}}$ is true, and 2. Let us assume that the ${{S}_{k}}$ be true for some integer $ k $. Then, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $ n\ge 3$. For, ${{S}_{3}}$ put $ n=3$ in the equation; $\left( {{3}^{2}} \right)>\left( \left( 2\times 3 \right)+1 \right)$ And simplify, we get: $9>7$ Then, one has satisfied the first equation Going to the second condition, assume that $ S\left( n \right)$ is true for $ n=k $ and prove that the statement is true for $ n=k+1$. ${{S}_{k}}:{{k}^{2}}>2k+1$ (I) And ${{S}_{k+1}}:{{k}^{2}}>2$ ${{S}_{k+1}}:{{\left( k+1 \right)}^{2}}>2\left( k+1 \right)+1$ And simplify the terms $\begin{align} & {{k}^{2}}+1+2k>2k+2+1 \\ & {{k}^{2}}>2 \\ \end{align}$ It is true for all the values of $ k>3$ So, this inequality is the same as the previous one which means it has satisfied the second condition of mathematical induction. Thus, by the method of mathematical induction, it is proved that ${{n}^{2}}>2n+1$ for $ n\ge 3$.
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