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: 29

Answer

$\left(\begin{array}{l} 20\\ 0 \end{array}\right)x^{20}(2y)^0=x^{20}$ $\left(\begin{array}{l} 20\\ 1 \end{array}\right)x^{19}(2y)^1=40x^{19}y$ $\left(\begin{array}{l} 20\\ 2\end{array}\right)x^{18}(2y)^{2}=760x^{18}y^{2}$

Work Step by Step

We find the first 3 terms in the expansion of $(x+2y)^{20}$ by using Binomial Theorem coefficients: $\left(\begin{array}{l} 20\\ 0 \end{array}\right)x^{20}(2y)^0=1*x^{20}(2y)^0=x^{20}$ $\left(\begin{array}{l} 20\\ 1 \end{array}\right)x^{19}(2y)^1=20*2x^{19}y=40x^{19}y$ $\left(\begin{array}{l} 20\\ 2\end{array}\right)x^{18}(2y)^{2}=190*4x^{18}y^2=760x^{18}y^{2}$
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.