Answer
See proof
Work Step by Step
We will rewrite
$S_n=\dfrac{a_1+a_n}{2}n$
taking advantage of the fact that
$a_n=a_1+d(n-1).$
So
$S_n=\dfrac{a_1+a_1+d(n-1)}{2}=\dfrac{n}{2}[2a_1+(n-1)d].$
Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
by Bittinger, Marvin L.; Ellenbogen, David J.; Johnson, Barbara L.
Published by
Pearson
ISBN 10:
0-32184-874-8
ISBN 13:
978-0-32184-874-1
Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.2 Arithmetic Sequences and Series - 14.2 Exercise Set - Page 903: 58
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