Answer
$$ \sum_{k=0} ^5 C(5,k) = 2^5=\boxed{32}$$
Work Step by Step
By the binomial theorem, we know that $ \sum_{k=0} ^n C(n,k) = 2^n$
Applying this theorem we get
$$ \sum_{k=0} ^5 C(5,k) = 2^5=\boxed{32}$$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter 8, Sequences and Series - Chapter 8 Review - Exercises - Page 641: 75
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