old-exam1-solns

# old-exam1-solns - Illinois Institute of Technology...

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Unformatted text preview: Illinois Institute of Technology Department of Computer Science First Examination Solutions CS 330 Discrete Mathematics Spring, 2007 1:50 PM–3:05 PM , Wednesday, February 21, 2007 238 Stuart Building Print your name and student ID, neatly in the space provided below; print your name at the upper right corner of every page. Please print legibly. Name: Student ID: This is an open book exam. You are permitted to use the textbook, any class handouts, anything posted on the web page, any of your own assignments, anything in your own handwriting, and a calculator. Foreign students may use a dictionary. Nothing else is permitted : No calculators, laptops, cell phones, etc.! Do all five problems in this booklet. All problems are equally weighted, so do not spend too much time on any one question. Show your work! You will not get partial credit if the grader cannot figure out how you arrived at your answer. Question Points Score Grader 1 20 2 20 3 20 4 20 5 20 Total 100 CS 330 First Exam—Spring, 2007 2 Name: 1. Mathematical Induction. 2+2+16 points Prove by induction that n 2 + n- 1 X k =1 (- 1) n- k k 2 = n + 1 2 ! . Sol: Let P ( n ) be the given proposition with n ≥ 1. We induct on n . Basis Step: LHS = 1 2 + ∑ 1- 1 k =1 (- 1) 1- k = 1; RHS = ( 1+1 2 ) = 1 Since LHS=RHS, P(1) is true Induction Hypothesis : Assume that the P ( m ) is true for a positive integer m . ⇒ P ( m ) : m 2 + ∑ m- 1 k =1 (- 1) m- k k 2 = ( m +1 2 ) ——- (1) Inductive Step: To show that ∀ P ( m ) → P ( m + 1) is true for m ≥ 1, we must show that if P ( m ) is true (the inductive hypothesis), then P ( m + 1) is true. Note that P ( m + 1) is the following statement: ( m + 1) 2 + ∑ m +1- 1 k =1 (- 1) m +1- k k 2 = ( m +1+1 2 ) ———- (2) Consider LHS of (2) = ( m + 1) 2 + ∑ m +1- 1 k =1 (- 1) m +1- k k 2 = ( m + 1) 2- ∑ m k =1 (- 1) m- k k 2 = ( m + 1) 2- ((- 1) m- m m 2 + ∑ m...
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old-exam1-solns - Illinois Institute of Technology...

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