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 - Page 636: 42



Work Step by Step

We find the term without $x$ in the expansion of $(8x+\frac{1}{2x})^{8}$ by using Binomial Theorem coefficients. To figure out when this occurs, we need to find out which term has no $x$ (e.g. has $x^0$). We know the general form of the terms is: $\displaystyle \left(\begin{array}{l} 8\\r\end{array}\right)(8x)^{r}(\frac{1}{2x})^{8-r}=\left(\begin{array}{l} 8\\r\end{array}\right)\frac{8^r}{2^{8-r}}*\frac{x^r}{x^{8-r}}$ We need to have the power of $x$ be zero, so: $\frac{x^r}{x^{8-r}}=x^0$ $x^{r-(8-r)}=x^{0}$ $x^{2r-8}=x^{0}$ $2r-8=0$ $r=8/2=4$ Thus we find the 4th term: $\displaystyle \left(\begin{array}{l} 8\\4\end{array}\right)(8x)^{4}(\frac{1}{2x})^{4}=70\frac{8^{4}x^4}{2^{4}x^4}=70*256=17920$
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.