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: 30

Answer

$\left(\begin{array}{l} 30\\ 0 \end{array}\right)(x^{1/2})^{30}(1)^0=x^{15}$ $\left(\begin{array}{l} 30\\ 1 \end{array}\right)(x^{1/2})^{29}(1)^1=30x^{29/2}$ $\left(\begin{array}{l}30\\2\end{array}\right)(x^{1/2})^{28}(1)^{2}=435x^{14}$ $\left(\begin{array}{l}30\\3\end{array}\right)(x^{1/2})^{27}(1)^{3}=4060x^{27/2}$

Work Step by Step

We find the first 4 terms in the expansion of $(x^{1/2}+1)^{30}$ by using Binomial Theorem coefficients: $\left(\begin{array}{l} 30\\ 0 \end{array}\right)(x^{1/2})^{30}(1)^0=x^{15}*1=x^{15}$ $\left(\begin{array}{l} 30\\ 1 \end{array}\right)(x^{1/2})^{29}(1)^1=30x^{29/2}$ $\left(\begin{array}{l}30\\2\end{array}\right)(x^{1/2})^{28}(1)^{2}=435x^{14}$ $\left(\begin{array}{l}30\\3\end{array}\right)(x^{1/2})^{27}(1)^{3}=4060x^{27/2}$
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