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: 43


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Work Step by Step

The expression is given below: ${{S}_{n}}=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\cdots +\frac{1}{2n\left( n+1 \right)}$ And the above sum can be written in a general from ${{S}_{n}}=\sum\limits_{n=1}^{n}{\frac{1}{2n\left( n+1 \right)}}$ It can be simplified as $\begin{align} & {{S}_{n}}=\frac{1}{2}\sum\limits_{n=1}^{n}{\frac{\left( 2n+2-2n \right)}{2n\left( n+1 \right)}} \\ & {{S}_{n}}=\frac{1}{2}\sum\limits_{n=1}^{n}{\frac{2\left( n+1 \right)-2n}{2n\left( n+1 \right)}} \\ \end{align}$ ${{S}_{n}}=\frac{1}{2}\sum\limits_{n=1}^{n}{\left( \frac{1}{n}-\frac{1}{n+1} \right)}$ and by putting the values of n in the above equation $\begin{align} & {{S}_{n}}=\frac{1}{2}\left( \left( 1-\frac{1}{2} \right)+\left( \frac{1}{2}-\frac{1}{3} \right)+\left( \frac{1}{3}-\frac{1}{4} \right)\ldots \left( \frac{1}{n}-\frac{1}{n+1} \right) \right) \\ & {{S}_{n}}=\frac{1}{2}\left( 1-\frac{1}{n+1} \right) \\ & {{S}_{n}}=\frac{n}{2\left( n+1 \right)} \\ \end{align}$ $(I)$ Now, prove the conjectural formula by mathematical induction Let ${{S}_{n}}$ be a statement involving the positive integer n. If 1. ${{S}_{1}}$ 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 $. For ${{S}_{1}}$ put $ n=1$ in the above equation, one gets ${{S}_{1}}=\frac{1}{4}$, It is the same as the first term, thus the first equation is satisfied. Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction. ${{S}_{k}}=\frac{k}{2\left( k+1 \right)}$ $(II)$ And for $ n=k+1$ ${{S}_{k+1}}=\frac{\left( k+1 \right)}{2\left( k+2 \right)}$ Now, one has $\begin{align} & {{S}_{k+1}}={{S}_{k}}+\left( \frac{1}{2(k+1)(k+2)} \right) \\ & {{S}_{k+1}}=\left( \frac{k}{2(k+1)} \right)+\left( \frac{1}{2(k+1)(k+2)} \right) \\ \end{align}$ And simplify ${{S}_{k+1}}=\frac{\left( k+1 \right)}{2\left( k+2 \right)}$ So, the inequality for $ k+1$ is the same as equation (I), which means it has satisfied the second condition of mathematical induction. Thus, the expression ${{S}_{n}}=\frac{n}{2\left( n+1 \right)}$ is true for all positive integers $ n $. And the value are ${{S}_{1}}=\frac{1}{4}$, ${{S}_{2}}=\frac{1}{3}$, ${{S}_{3}}=\frac{3}{8}$, and ${{S}_{5}}=\frac{5}{12}$
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