Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.1 An Overview - 8.1 Exercises: 16

Answer

$0, 3, 10,21$

Work Step by Step

$a_n = 2n^2 - 3n + 1$ To find the first four terms, we need to find $a_1, a_2, a_3$, and $a_4$. $a_1 = 2(1)^2 - 3(1) + 1 = 0$ $a_2 = 2(2)^2 - 3(2) + 1 = 3$ $a_3 = 2(3)^2 - 3(3) + 1 = 10$ $a_4 = 2(4)^2 - 3(4) + 1 = 21$
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