Stitz-Zeager_College_Algebra_e-book

Note that the terminal side of lies radians short of

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Unformatted text preview: Dividing, we get 7! 5040 5! = 120 = 42. While this is correct, we note that we could have saved ourselves some of time had we proceeded as follows 7! 7·6·5·4·3·2·1 7·6·5·4·3·2·1 ¡¡¡¡¡ = = = 7 · 6 = 42 5! 5·4·3·2·1 5·4·3·2·1 ¡¡¡¡¡ In fact, should we want to fully exploit the recursive nature of the factorial, we can write 5! 7 · 6 · 5! 7 · 6 · 7! = = = 42 5! 5! 5! (c) Keeping in mind the lesson we learned from the previous problem, we have $ 1000 · 999 · 998! 1000 · 999 · $$ 998! 999000 1000! = = = = 499500 $ $ · 2! $ 998! 2! 998! · 2! 998! 2 (d) This problem continues the theme which we have seen in the previous two problems. We first note that since k + 2 is larger than k − 1, (k + 2)! contains all of the factors of (k − 1)! and as a result we can get the (k − 1)! to cancel from the denominator. To see this, we begin by writing out (k + 2)! starting with (k + 2) and multiplying it by the numbers which precede it until we reach (k − 1): (k + 2)! = (k + 2)(k + 1)(k )(k − 1)!. As a result, we have (k + 2)! (k + 2)(k + 1)(k )(k − 1)! (k + 2)(k + 1)(k )$$$$ (...
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