225_17_inductive_dfn

# 225_17_inductive_dfn - (1)=1,T(n)=1 T(n1)is 1 2 3 n Theorem:(n 1/2 Proof[Basis]T(1)=1and1(1 1/2=1asrequired[Inductionstep]Assumethat1 2 n1 n=n(n

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What is wrong with my induction proof? In a drunken haze I decided that the solution to the recurrence T(1)=1, T(n)= 1  + T(n-1) is  1 + 2 + 3 + … + n. Theorem: The solution to the recurrence is n(n+1)/2. Proof. [Basis] T(1)=1 and 1 *(1+1)/2 = 1 as required. [Induction step] Assume that 1 + 2 + … + n-1 + n = n(n+1)/2. We want to prove that 1 + 2 + … + n-1 + n + (n+1) = (n+1)(n+2)/2 = (n 2  +3n  +2)/2. By induction, 1 + 2 + … + n= n(n+1)/2. So 1 + 2 + . . + n + (n+1)=  n(n+1)/2 + (n+1). Simplifying: (n 2  + n + 2n + 2)/2= (n 2  +3n +2)/2 as required.

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Old midterms are posted. Midterm tutorial: Sunday Oct. 16, 1pm, ECS 116. Monday Oct. 17: No office hours. If you need help you may attend one or  both of the labs for CSC 225. B01: Monday, 12:00-12:50 pm in DSB C 112. B02: Monday, 1:00-1:50 pm,  ECS 116. Midterm is in class on Tues. Oct. 18. Read through the exam carefully and choose the easiest questions to answer  first to make sure you get at least 40% on the midterm (a requirement for  writing the final exam).
Midterm material: You are responsible for all assignment material and all material covered in class. Mathematics: Induction, recurrence relations, definitions of O, Ω, and Θ (you don’t have to know them word for word as long as your definition is mathematically equivalent to mine) and how to use them, formulas for my 3 favorite sums, lower bounding and upper bounding, logarithms base 2, floor and ceiling functions.

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## This note was uploaded on 01/15/2012 for the course CSC 225 taught by Professor Valerieking during the Spring '10 term at University of Victoria.

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225_17_inductive_dfn - (1)=1,T(n)=1 T(n1)is 1 2 3 n Theorem:(n 1/2 Proof[Basis]T(1)=1and1(1 1/2=1asrequired[Inductionstep]Assumethat1 2 n1 n=n(n

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