CIS 2910 Lecture 3: Lab 03 Solutions

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1 induction: prove by induction that (where n > 0) (a) n 1. [bc]: n = 1; lhs = 1 ((1)+1)! {z use rhs in [ih] (k + 1)! = (k+2)! (k+2)+(k+1) (k+2)! but n = k + 1 (n+1)! 1 (n+1)! 1 (c) 4n < (n2 7) for n 6. [bc]: n = 6; lhs = 4(6) = 24 rhs = (6)2 7 = 29 lhs < rhs x. 4k < (k2 7) for k 6. |{z} use rhs in [ih] since k 6; 4 < 2k + 1 is true k2 + 2k + 1 7 (k + 1)2 7 but n = k + 1; n2 7. = 24 rhs = 24 = 16 lhs < rhs x. [ih]: assume true for n = k; k! [is]: try n = k + 1; (k + 1)! > k(2k) + (2k) since k 4 , k is replaced by 1 > (2k) + (2k)

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