Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Problems - Page 76: 18


Proved below.

Work Step by Step

1 + $(2n-1)$ is $2n$. So is $3 + (2n - 3)$. And so on. This occurs with each of the $n/2$ pairings of numbers. Thus after rearranging, the sum $1 + 3 + 5 + \cdots + (2n - 1)$ is $2n\cdot (n/2) = n^2$. This can be proved more formally by induction.
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