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: 36

Answer

$a_{n}=\frac{n+2}{n+3}$ Or: $a_{n}=\frac{3+(n-1)}{4+(n-1)}$

Work Step by Step

We are given: $\frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}$ ... We notice that the numerator and denominator both increase by $1$ between terms, but start with different values. We find the pattern: $a_1=\frac{1+2}{1+3}$ $a_2=\frac{2+2}{2+3}$ $a_3=\frac{3+2}{3+3}$ $a_4=\frac{4+2}{4+3}$ Therefore: $a_{n}=\frac{n+2}{n+3}$ Or: $a_{n}=\frac{3+(n-1)}{4+(n-1)}$
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.