## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 8, Sequences and Series - Section 8.6 - The Binomial Theorem - 8.6 Exercises - Page 636: 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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