Unformatted text preview: + 1) 23, since k + 1 > 1. Case 2 ( k + 1 ≥ 8): Now T ( k + 1) = 4 T ( b k +1 2 c ) + 7 by the deﬁnition < 4(7( b k +1 2 c ) 23) + 7 by IH since b k +1 2 c ≤ k +1 2 ≤ k when k ≥ 1 ≤ 4(7( k +1 2 ) 23) + 7 since b k +1 2 c ≤ k +1 2 = 7( k + 1) 212 + 7 < 7( k + 1) 23 which is what we needed to show 1...
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 Spring '10
 fleck
 Mathematical Induction, Recursion, Inductive Reasoning, Mathematical logic, base case, strong inductive hypothesis

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