College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 8 - Sequences, Induction, and Probability - Exercise Set 8.2 - Page 727: 84

Answer

The sum of the first n positive odd integers 1 + 3 + 5 + . . . . . . . . . +(2n - 1) = $n^{2}$ Proof- $1^{st}$ term $a_{1}$ = 1 Common difference = 2 Last term $a_{L}$ = (2n - 1) The sum $S_{n}$ = $\frac{n}{2}$(first term + last term) = $\frac{n}{2}$($a_{1}$ + $a_{L}$) = $\frac{n}{2}$(1 + 2n - 1) = $\frac{n}{2}$(2n) =n$\times$n =$ n^{2}$

Work Step by Step

The sum of the first n positive odd integers 1 + 3 + 5 + . . . . . . . . . +(2n - 1) = $n^{2}$ Proof- $1^{st}$ term $a_{1}$ = 1 Common difference = 2 Last term $a_{L}$ = (2n - 1) The sum $S_{n}$ = $\frac{n}{2}$(first term + last term) = $\frac{n}{2}$($a_{1}$ + $a_{L}$) = $\frac{n}{2}$(1 + 2n - 1) = $\frac{n}{2}$(2n) =n$\times$n =$ n^{2}$
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