Adv Alegbra HW Solutions 2

Adv Alegbra HW Solutions 2 - n 0. 1.3 Show, for all n 1,...

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2 Solution. True. We have 0 = F 0 ,1 = F 1 ,1 = F 2 , and 2 = F 3 . Use the second form of induction with base steps n = 2 and n = 3 (verifying the inductive step will show why we choose these numbers). By the inductive hypothesis, n 2 F n 2 and n 1 F n 1 . Hence, 2 n 3 F n .Bu t n 2 n 3 for all n 3, as desired. (viii) If m and n are natural numbers, then ( mn ) != m ! n ! . Solution. False. If m = 2 = n , then ( mn ) != 24 and m ! n != 4. 1.2 (i) For any n 0 and any r ±= 1, prove that 1 + r + r 2 + r 3 +···+ r n = ( 1 r n + 1 )/( 1 r ). Solution. We use induction on n 1. When n = 1, both sides equal 1 + r . For the inductive step, note that [ 1 + r + r 2 + r 3 +···+ r n ]+ r n + 1 = ( 1 r n + 1 )/( 1 r ) + r n + 1 = 1 r n + 1 + ( 1 r ) r n + 1 1 r = 1 r n + 2 1 r . (ii) Prove that 1 + 2 + 2 2 +···+ 2 n = 2 n + 1 1 . Solution. This is the special case of the geometric series when r = 2; hence, the sum is ( 1 2 n + 1 )/( 1 2 ) = 2 n + 1 1. One can
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Unformatted text preview: n 0. 1.3 Show, for all n 1, that 10 n leaves remainder 1 after dividing by 9. Solution. This may be rephrased to say that there is an integer q n with 10 n = 9 q n + 1. If we dene q 1 = 1, then 10 = q 1 + 1, and so the base step is true. For the inductive step, there is an integer q n with 10 n + 1 = 10 10 n = 10 ( 9 q n + 1 ) = 90 q n + 10 = 9 ( 10 q n + 1 ) + 1 . Dene q n + 1 = 10 q n + 1, which is an integer. 1.4 Prove that if 0 a b , then a n b n for all n 0. Solution. Base step . a = 1 = b , and so a b . Inductive step . The inductive hypothesis is a n b n ....
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