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: 44


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Work Step by Step

The expression is given below: ${{S}_{n}}=\left( 1-\frac{1}{2} \right)\times \left( 1-\frac{1}{3} \right)\times \left( 1-\frac{1}{4} \right)........\left( 1-\frac{1}{\left( n+1 \right)} \right)$ And the above sum can be written in a general from ${{S}_{n}}=\prod\limits_{n=1}^{n}{\left( 1-\frac{1}{(n+1)} \right)}$ It can be simplified as $\begin{align} & {{S}_{n}}=\prod\limits_{n=1}^{n}{\left( 1-\frac{1}{(n+1)} \right)} \\ & {{S}_{n}}=\prod\limits_{n=1}^{n}{\left( \frac{n}{n+1} \right)} \\ \end{align}$ Substitute the values of n in the above equation ${{S}_{n}}=\left( \frac{1}{2} \right)\times \left( \frac{2}{3} \right)\times \left( \frac{3}{4} \right)......\left( \frac{n}{\left( n+1 \right)} \right)$ ${{S}_{n}}=\frac{1}{\left( n+1 \right)}$ $(I)$ To 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 ${{S}_{1}}=\frac{1}{2}$, 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. ${{S}_{k}}=\frac{1}{\left( k+1 \right)}$ $(II)$ Then, one has to prove ${{S}_{k+1}}$ is true ${{S}_{k+1}}=\frac{1}{k+2}$ Now, one has $\begin{align} & {{S}_{k+1}}={{S}_{k}}\times \left( \frac{k+1}{k+2} \right) \\ & {{S}_{k+1}}=\left( \frac{1}{k+1} \right)\times \left( \frac{k+1}{k+2} \right) \\ \end{align}$ And simplify ${{S}_{k+1}}=\frac{1}{k+2}$ This inequality for $ k+1$ is the same as equation (I); this means it has satisfied the second condition of mathematical induction. So, the expression ${{S}_{n}}=\left( 1-\frac{1}{2} \right)\times \left( 1-\frac{1}{3} \right)\times \left( 1-\frac{1}{4} \right)........\left( 1-\frac{1}{\left( n+1 \right)} \right)$ is true for all positive integers $ n $. Thus, the value are ${{S}_{1}}=\frac{1}{2}$, ${{S}_{2}}=\frac{1}{3}$, ${{S}_{3}}=\frac{1}{4}$, ${{S}_{4}}=\frac{1}{5}$ and ${{S}_{5}}=\frac{1}{6}$.
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