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: 24

Answer

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

Work Step by Step

We evaluate: $\displaystyle \left(\begin{array}{l} 5\\ 0 \end{array}\right)-\left(\begin{array}{l} 5\\ 1 \end{array}\right)+\left(\begin{array}{l} 5\\ 2 \end{array}\right)-\left(\begin{array}{l} 5\\ 3 \end{array}\right)+\left(\begin{array}{l} 5\\ 4 \end{array}\right)-\left(\begin{array}{l} 5\\ 5 \end{array}\right)=\frac{5!}{0!5!}-\frac{5!}{1!4!}+\frac{5!}{2!3!}-\frac{5!}{3!2!}+\frac{5!}{4!1!}-\frac{5!}{5!0!}=0$ Because all the terms cancel.
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