Answer
$n^3-3n^2+2n$
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, $=n^3-3n^2+2n$
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
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
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