answers2 - EECS 203 Fall 2010 Midterm Exam 2 (75 points)...

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1 EECS 203 Fall 2010 Midterm Exam 2 (75 points) Closed Book Closed Electronic Devices Closed Neighbor Two 8.5 x 11 pages are allowed as cheat sheets. Name: Sample Solution UM unique name: @umich.edu Discussion section: (circle) Benjamin(Fri 1:30-2:30) Pradeep(Mon 3:30-4:30) Michael(Mon 12:30-1:30) Benjamin(Fri 2:30-3:30) Pradeep(Fri 3:30-4:30) Michael(Mon 1:30-2:30) Kuldeep(Fri 12:30-1:30) Instructions: 1. Write clearly. If we cannot read your writing then it will be marked wrong. 2. You may use the blank pages as spillover space or for scratch work. 3. This course operates under the rules of the College of Engineering Honor Code. Your signature endorses the pledge below. After you finish your exam, please sign below: I have neither given nor received aid on this examination, nor have I concealed any violations of the Honor Code. Q 1 /10 Q 5 /10 Q 2 /10 Q 6 /05 Q 3 /10 Q 7 /10 Q 4 /10 Q8 /10 Total /75

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2 1) (10 points) Prove using mathematical induction that for every positive integer n: 1 + 4 + 7 + 10 + … + (3 n - 2) = 2 3 2 n n We prove using induction. Let P( n ) denote the predicate: 1 + 4 + 7 + 10 + … + (3 n - 2) = (3 n 2 n )/2 Basis Step: We prove P(1) L.H.S. of P(1) = 1 R.H.S. of P(1) = (3 × 1 2 – 1)/2 = 1 P(1) is true Inductive Step: We prove k > 0 (P( k ) P( k +1)) Let P( k ) be true for an arbitrary k > 0. That is, let: 1 + 4 + 7 + 10 + … + (3 k - 2) = (3 k 2 k )/2 ( inductive hypothesis ) L.H.S of P( k +1) = 1 + 4 + 7 + 10 + … + (3 k - 2) + (3( k +1) - 2) = (3 k 2 k )/2 + (3( k +1) - 2) ( using the inductive hypothesis ) = (3 k 2 + 5 k + 2)/2 R.H.S. of P( k +1) = (3( k +1) 2 – ( k +1))/2 = (3 k 2 + 5 k + 2)/2 P( k +1) is true.
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.

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answers2 - EECS 203 Fall 2010 Midterm Exam 2 (75 points)...

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