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 - Page 637: 51


Yes, the statement is true.

Work Step by Step

$\displaystyle \left(\begin{array}{l}n\\1\end{array}\right)=\frac{n!}{1!(n-1)!}=\frac{n(n-1)!}{1*(n-1)!}=\frac{n}{1}=n$ $\displaystyle \left(\begin{array}{l} n\\n-1\end{array}\right)=\frac{n!}{(n-1)!1!}=\frac{n(n-1)!}{(n-1)!*1}=\frac{n}{1}=n$ Thus we have shown that: $\left(\begin{array}{l}n\\1\end{array}\right)=\left(\begin{array}{l}n\\n-1\end{array}\right)=n$
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