Stitz-Zeager_College_Algebra_e-book

3 if sin 1 nd cos solution 1 when we substitute

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1! = 2 · 1 = 2, 3! = 3 · 2! = 3 · 2 · 1 = 6 and 4! = 4 · 3! = 4 · 3 · 2 · 1 = 24. Informally, n! = n · (n − 1) · (n − 2) · · · 2 · 1 with 0! = 1 as our ‘base case.’ Our first example familiarizes us with some of the basic computations involving factorials. Example 9.4.1. 1. Simplify the following expressions. (a) 3! 2! 0! (b) 7! 5! (c) 1000! 998! 2! (d) (k + 2)! ,k≥1 (k − 1)! 2. Prove n! > 3n for all n ≥ 7. Solution: 1. We keep in mind the mantra, “When in doubt, write it out!” as we simplify the following. (a) We have been programmed to react with alarm to the presence of a 0 in the denominator, but in this case 0! = 1, so the fraction is defined after all. As for the numerator, 3! = 3 · 2 · 1 = 6 and 2! = 2 · 1 = 2, so we have 3! 2! = (6)(2) = 12. 0! 1 1 It’s pretty much the reason Section 9.3 is in the book. 582 Sequences and the Binomial Theorem (b) We have 7! = 7 · 6 · 5 · 4 · 3 · 2 · 1 = 5040 while 5! = 5 · 4 · 3 · 2 · 1 = 120....
View Full Document

This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

Ask a homework question - tutors are online