College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises: 35

Answer

$a_{n}= \frac{2n-1}{n^{2}}$ Or: $a_{n}= \frac{1+2(n-1)}{n^{2}}$

Work Step by Step

We are given: $\frac{1}{1}, \frac{3}{4}, \frac{5}{9}, \frac{7}{16}, \frac{9}{25}$ ... We notice that the numerators increase by $2$ between terms. We also see that the denominators are powers of integers ($1^2$, $2^2$, $3^2$, etc). We find the pattern: $a_{1}= \frac{2(1)-1}{1^{2}}$ $a_{2}= \frac{2(2)-1}{2^{2}}$ $a_{3}= \frac{2(3)-1}{3^{2}}$ $a_{4}= \frac{2(4)-1}{4^{2}}$ $a_{5}= \frac{2(5)-1}{5^{2}}$ Therefore: $a_{n}= \frac{2n-1}{n^{2}}$ Or: $a_{n}= \frac{1+2(n-1)}{n^{2}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.