Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.1 Sequences - 11.1 Assess Your Understanding - Page 827: 62


$\sum_{k=1}^{12} (2k-1)$

Work Step by Step

We need to compute summation formula for the sequence $1+3+5+7+.......+[2(12)-1]$ When $k=1$, the first term is $1=(2)(1)-1$ and for $k=2$ , the second term is $3=(2)(2)-1$. This information indicates the $k^{th}$ term as $(2k-1) $. So, we write the the summation formula for $k^{th}$ term as follows: $1+3+5+7+.......+[2(12)-1]= \sum_{k=1}^{12} (2k-1)$
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