Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Mid-Chapter Check Point - Page 858: 12



Work Step by Step

since,we have $\sum_{i=1}^{50}(3i-2)$ Here, terms are: {$1,4,7,10,...,148$}; which is an arithmetic sequence. Their sum is $S_n=\dfrac{n}{2}(a_1+a_n)$ $=\dfrac{50}{2}(1+148)$ or, $=25(149)$ Hence, $=3725$
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