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 - Page 639: 4

Answer

$a_1=1$ $a_2=3$ $a_3=6$ $a_4=10$ $a_{10}=55$

Work Step by Step

We are given: $a_{n}= \frac{n(n+1)}{2}$ We evaluate: $a_1= \frac{1(1+1)}{2}=\frac{1*2}{2}=1$ $a_2= \frac{2(2+1)}{2}=\frac{2*3}{2}=3$ $a_3= \frac{3(3+1)}{2}=\frac{3*4}{2}=6$ $a_4= \frac{4(4+1)}{2}=\frac{4*5}{2}=10$ $a_{10}= \frac{10(10+1)}{2}=\frac{10*11}{2}=55$
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