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: 1

Answer

The value of ${{S}_{1}}=1$, ${{S}_{2}}=4$ and ${{S}_{3}}=9$, and also the statement is true.

Work Step by Step

We consider the statement: ${{S}_{n}}:1+3+5+\cdots +\left( 2n-1 \right)={{n}^{2}}$ For ${{S}_{1}}$ we has ${{S}_{1}}:1={{1}^{2}}$ Therefore, the above statement is true for $ n=1$. For ${{S}_{2}}$ one has ${{S}_{2}}:1+3={{2}^{2}}$ Therefore, the above statement is true for $ n=2$. For ${{S}_{3}}$ one has ${{S}_{3}}:1+3+5={{3}^{2}}$ Therefore, the above statement is true for $ n=3$. Thus, the statement holds. Thus the values are ${{S}_{1}}=1,\,\,{{S}_{2}}=4$, and ${{S}_{3}}=9$. And the statement is true.
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