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: 23

Answer

$\left(\begin{array}{l} 5\\ 0 \end{array}\right)+\left(\begin{array}{l} 5\\ 1 \end{array}\right)+\left(\begin{array}{l} 5\\ 2 \end{array}\right)+\left(\begin{array}{l} 5\\ 3 \end{array}\right)+\left(\begin{array}{l} 5\\ 4 \end{array}\right)+\left(\begin{array}{l} 5\\ 5 \end{array}\right)=32$

Work Step by Step

We evaluate: $\left(\begin{array}{l} 5\\ 0 \end{array}\right)+\left(\begin{array}{l} 5\\ 1 \end{array}\right)+\left(\begin{array}{l} 5\\ 2 \end{array}\right)+\left(\begin{array}{l} 5\\ 3 \end{array}\right)+\left(\begin{array}{l} 5\\ 4 \end{array}\right)+\left(\begin{array}{l} 5\\ 5 \end{array}\right)$ We notice that this represents an expanded 5th power binomial. Instead of evaluating each term, we simply reverse the expansion into the original binomial: $=(1+1)^{5}=2^{5}=32$
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