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: 40

Answer

$3520\sqrt{2}y^{3}$

Work Step by Step

We find the term with $y^3$ in the expansion of $(\sqrt{2}+y)^{12}$ by using Binomial Theorem coefficients: $\left(\begin{array}{l} 12\\9\end{array}\right)(\sqrt{2})^{9}y^{3}=220*(\sqrt{2})^8\sqrt{2}y^3=220*16\sqrt{2}y^3=3520\sqrt{2}y^{3}$
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