College Algebra 7th Edition

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

Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises: 27

Answer

$\displaystyle (1+ \frac{1}{x})^{6}=1+\frac{6}{x}+\frac{15}{x^{2}}+\frac{20}{x^{3}}+\frac{15}{x^{4}}+\frac{6}{x^{5}}+\frac{1}{x^{6}}$ Or written with negative exponents: $\displaystyle (1+ \frac{1}{x})^{6}=1+6x^{-1}+15x^{-2}+20x^{-3}+15x^{-4}+6x^{-5}+x^{-6}$

Work Step by Step

We expand using the Binomial Theorem: $\displaystyle (1+ \frac{1}{x})^{6}=\left(\begin{array}{l} 6\\ 0 \end{array}\right)1^{6}+\left(\begin{array}{l} 6\\ 1 \end{array}\right)1^{5}(\frac{1}{x})^1+\left(\begin{array}{l} 6\\ 2 \end{array}\right)1^{4}(\frac{1}{x})^{2}+\left(\begin{array}{l} 6\\ 3 \end{array}\right)1^{3}(\frac{1}{x})^{3}+\left(\begin{array}{l} 6\\ 4 \end{array}\right)1^{2}(\frac{1}{x})^{4}+\left(\begin{array}{l} 6\\ 5 \end{array}\right)1^{1}(\frac{1}{x})^{5}+\left(\begin{array}{l} 6\\ 6 \end{array}\right)(\frac{1}{x})^{6}=1+\frac{6}{x}+\frac{15}{x^{2}}+\frac{20}{x^{3}}+\frac{15}{x^{4}}+\frac{6}{x^{5}}+\frac{1}{x^{6}}$ Or written with negative exponents: $=1+6x^{-1}+15x^{-2}+20x^{-3}+15x^{-4}+6x^{-5}+x^{-6}$
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.