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 725: 42

Answer

6480

Work Step by Step

Method 1- - Sum of first 80 positive even integers = 2 + 4 + 6 + 8 + 10 . . . . .+ 160 First term $a_{1}$ = 2 Common difference d = 2 $60^{th}$ term = 160 Sum of n terms = $\frac{n}{2}$($a_{1}$ + $a_{n}$) Sum of first 80 positive even integers = $\frac{80}{2}$(2 + 160) = 40$\times$162 = 6480 Method 2- - 2 + 4 + 6 + 8 + 10 . . . . .+ 160 = 2(1 + 2 + 3 + 4 + 5 +......+80) First find sum of (1 + 2 + 3 + 4 + 5 +......+80) and then multiply the sum by 2 First term of series $a_{1}$ = 1 Common difference in series d = 1 $80^{th}$ term = 80 Sum of n terms = $\frac{n}{2}$($a_{1}$ + $a_{n}$) Sum of first 80 natural numbers= $\frac{80}{2}$(1 + 80) = 40$\times$81 = 3240 2 + 4 + 6 + 8 + 10 . . . . .+ 160 = 2(1 + 2 + 3 + 4 + 5 +......+80) = 2 $\times$ 3240 = 6480
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