Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.1 Sequences and Series - 14.1 Exercise Set - Page 895: 87

Answer

$1,3,13,63,313,1563$.

Work Step by Step

${{a}_{n+1}}=5{{a}_{n}}-2$ ${{a}_{1}}=1$ $\begin{align} & {{a}_{n+1}}=5{{a}_{n}}-2 \\ & {{a}_{1+1}}=5{{a}_{1}}-2 \\ & {{a}_{2}}=5\cdot 1-2 \\ & =3 \end{align}$ $\begin{align} & {{a}_{n+1}}=5{{a}_{n}}-2 \\ & {{a}_{2+1}}=5{{a}_{2}}-2 \\ & {{a}_{3}}=5\cdot 3-2 \\ & =13 \end{align}$ $\begin{align} & {{a}_{n+1}}=5{{a}_{n}}-2 \\ & {{a}_{3+1}}=5{{a}_{3}}-2 \\ & {{a}_{4}}=5\cdot 13-2 \\ & =63 \end{align}$ $\begin{align} & {{a}_{n+1}}=5{{a}_{n}}-2 \\ & {{a}_{4+1}}=5{{a}_{4}}-2 \\ & {{a}_{5}}=5\cdot 63-2 \\ & =313 \end{align}$ $\begin{align} & {{a}_{n+1}}=5{{a}_{n}}-2 \\ & {{a}_{5+1}}=5{{a}_{5}}-2 \\ & {{a}_{6}}=5\cdot 313-2 \\ & =1563 \end{align}$ Therefore, $\begin{align} & {{a}_{1}}=1 \\ & {{a}_{2}}=3 \\ & {{a}_{3}}=13 \\ & \\ \end{align}$ And, $\begin{align} & {{a}_{4}}=63 \\ & {{a}_{5}}=313 \\ & {{a}_{6}}=1563 \\ \end{align}$ Thus, the first six terms are: $1,3,13,63,313,1563$.
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.