College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Chapter 8 Review - Exercises: 6

Answer

$a_{1}=1$ $a_{2}= 3$ $a_{3}=6$ $a_{4}=10$ $a_{10}=55$

Work Step by Step

We are given: $a_{n}=\left(\begin{array}{l}n+1\\2\end{array}\right)$ We evaluate: $a_{1}=\left(\begin{array}{l}1+1\\2\end{array}\right)=\frac{2!}{2!0!}=1$ $a_{2}= \left(\begin{array}{l}2+1\\2\end{array}\right)=\frac{3!}{2!1!}=3$ $a_{3}= \left(\begin{array}{l}3+1\\2\end{array}\right)=\frac{4!}{2!2!}=6$ $a_{4}= \left(\begin{array}{l}4+1\\2\end{array}\right)=\frac{5!}{2!3!}=10$ $a_{10}= \left(\begin{array}{l}10+1\\2\end{array}\right)=\frac{11!}{2!9!}=55$
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