Answer
$d(n)=32n-16$
Work Step by Step
From part (a), we have:
$d(n)=D(n)-D(n-1)$
Thus, we have
$d(n)=16n^2 -16(n-1)^2$
When we plug in $n=1,2,3,4$:
This gives: $16, 48,80,112$
$d(n)=16n^2 -16(n^2-2n+1)$
Thus, $d(n)=32n-16$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
Published by
McDougal Littell
ISBN 10:
0618595414
ISBN 13:
978-0-61859-541-9
Chapter 12 Sequences and Series - 12.2 Analyze Arithmetic Sequences and Series - 12.2 Exercises - Problem Solving - Page 808: 66b
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
