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 1085: 13


See the explanation below.

Work Step by Step

We use mathematical induction as follows: Statement ${{S}_{1}}$ is; $1=\left( 2-1 \right)$ After simplifying on the right, we obtain ${{n}^{2}}=1$. This true statement shows that ${{S}_{1}}$ is true. Suppose ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression, ${{S}_{k}}=1+3+5+....+\left( 2k-1 \right)={{k}^{2}}$ Adding $\left( 2\left( k+1 \right)-1 \right)$ on both sides as: $\begin{align} & 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{k}^{2}}+\left( 2\left( k+1 \right)-1 \right) \\ & 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{k}^{2}}+\left( 2k+2-1 \right) \\ & 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{k}^{2}}+2k+1 \\ & 1+3+5+\ldots +\left( 2\left( k+1 \right)-1 \right)={{\left( k+1 \right)}^{2}} \end{align}$ Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=1+3+5+\ldots +\left( 2n-1 \right)={{n}^{2}}$ holds true
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