hw5_key - COT3100 Summer2001 Assignment#5(Solution Assigned...

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COT3100 Summer’2001 Assignment #5 (Solution) Assigned: 07/05, due: 07/19 in recitation Total 100 points. 1. (8pts) Prove by induction that for every nonnegative integer n 0, 2 0 +2 1 +… 2 n =2 n +1 - 1. Basis . n =0: LHS=2 0 =1; RHS=2 0+1 - 1 = 2 - 1 =1. So, LHS=RHS. IH . Assume that for n = k , where k is some integer k 0, the equality holds. In other words assume that for some k 0 2 0 +2 1 +…2 k =2 k +1 - 1. IS . We need to prove that the equality holds for n = k +1, i.e. 2 0 +2 1 +…2 k + 2 k +1 = 2 ( k +1)+1 - 1. 2 0 +2 1 +…2 k + 2 k +1 = (2 k +1 - 1) + 2 k +1 (by IH) = 2 2 k +1 - 1 = 2 k +2 - 1 So, induction step is proved. By induction principle the summation formula is true for any n 0. 2. (9pts) A sequence of numbers a n is defined recursively as follows: a 0 =0; a n +1 =2 a n + n , for n 1 Prove that a n =2 n - n - 1. Basis . We need to check that the given formula is correct for n = 0. This formula gives a 0 =2 0 - 0 - 1 = 1 - 1 = 0, in agreement with a 0 = 0 by recursive definition. IH . Assume that the formula is correct for n = k , where k is some integer k 0, i.e. we assume that a k = 2 k - k - 1. IS We need to prove that the formula is correct for n = k +1, i.e. that a k +1 =2 k +1 - ( k +1) - 1 = = 2 k +1 - k - 2. But by recursive definition we have that a k +1 =2 a k + k . By substitution a k =2 k - k - 1 from IH we get a k +1 =2 (2 k - k - 1) + k = 2 k +1 - 2 k - 2 + k = 2 k +1 - k - 2. So, induction step is proved. By induction principle the formula a n =2 n - n - 1 is correct for any n 0. 3. (15pts) Use induction on n to prove the following inequality for all positive integers n : 6 ) 1 4 )( 1 ( 2 1 - + = n n n i n i Basis . n =1 : LHS = = = = = 1 1 1 1 i i i , RHS = 1 6 ) 1 4 )( 1 1 ( 1 = - + , so the inequality LHS RHS holds.

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