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 - Section 11.1 - Sequences and Summation Notation - Exercise Set - Page 831: 91



Work Step by Step

Since,we have $\dfrac{n!}{(n-3)!}$ Thus, $\dfrac{n!}{(n-3)!}=\dfrac{n \cdot (n-1) \cdot (n-2) \cdot (n-3) !}{(n-3)!}$ or, $=n(n-1)(n-2)$ or, $=n(n^2-3n+2)$ or, $=n^3-3n^2+2n$
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