Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 10 - Section 10.1 - Sequences and Summation Notation - Exercise Set - Page 1048: 41



Work Step by Step

Here, we have $\sum_{i=1}^5 \dfrac{i!}{(i-1)!}= \dfrac{1!}{(1-1)!}+ \dfrac{2!}{(2-1)!}+ \dfrac{3!}{(3-1)!}+ \dfrac{4!}{(4-1)!}+ \dfrac{5!}{(5-1)!}$ $=\dfrac{1}{1}+\dfrac{2}{1}+\dfrac{6}{2}+\dfrac{24}{6}+\dfrac{120}{24}$ $=1+2+3+4+5$ $=15$
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