Precalculus (6th Edition) Blitzer

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

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Concept and Vocabulary Check - Page 1048: 4


$\underline{{{a}_{2}}=\frac{{{\left( -1 \right)}^{2}}}{{{4}^{2}}-1}=\frac{1}{15}}$.

Work Step by Step

Put $n=2$ to get the second term of the sequence given. Now, to get the third term, put $n=3$ and so on. We have the general form of the sequence: ${{a}_{n}}=\frac{{{\left( -1 \right)}^{n}}}{{{4}^{n}}-1}$. So to find the second term, put $n=2$: $\begin{align} & {{a}_{2}}=\frac{{{\left( -1 \right)}^{2}}}{{{4}^{2}}-1} \\ & =\frac{1}{16-1} \\ & =\frac{1}{15} \end{align}$
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