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: 42



Work Step by Step

Here, we have $\sum_{i=1}^5 \dfrac{(i+2)!}{i!}= \dfrac{(1+2)!}{1!}+ \dfrac{(2+2)!}{2!}+ \dfrac{(3+2)!}{3!}+ \dfrac{(4+2)!}{4!}+ \dfrac{(5+2)!}{5!}$ $=\dfrac{3!}{1!}+\dfrac{4!}{2!}+\dfrac{5!}{3!}+\dfrac{6!}{4!}+\dfrac{7!}{5!}$ $=\dfrac{6}{1}+\dfrac{24}{2}+\dfrac{120}{6}+\dfrac{720}{24}+\dfrac{5040}{120}$ $=110$
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