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MIT6_042JS10_lec09_sol

MIT6_042JS10_lec09_sol - Massachusetts Institute of...

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Massachusetts Institute of Technology 6.042J/18.062J, Spring ’10 : Mathematics for Computer Science February 22 Prof. Albert R. Meyer revised February 18, 2010, 15 minutes Solutions to In-Class Problems Week 4, Mon. Problem 1. Prove by induction: 1 1 1 1 1 + 4 + 9 + · · · + n 2 < 2 n , (1) for all n > 1 . Solution. Proof. (By Induction). The induction hypothesis is P ( n ) is the inequality ( 1 ). Base Case : ( n = 2 ). The LHS of( 1 ) in this case is 1 + 1 / 4 and the RHS is 2 1 / 2 . Since LHS = 5 / 4 < 6 / 4 = 3 / 2 = RHS, inequality ( 1 ) holds, and P (2) is proved. Inductive Step : Let n be any natural number greater than 1 , and assume P ( n ) in order to prove P ( n + 1) . That is, we assume ( 1 ) Adding 1 / ( n + 1) 2 to both sides of this inequality yields 1 1 1 1 1 1 + 4 + · · · + n 2 + ( n + 1) 2 < 2 n + ( n + 1) 2 1 1 = 2 n ( n + 1) 2 n 2 + 2 n + 1 n = 2 n ( n + 1) 2 n 2 + n 1 = 2 n ( n + 1) 2 n ( n + 1) 2 1 1 = 2 n + 1 n ( n + 1) 2 1 . < 2 n + 1 So we have proved P ( n + 1) . Problem 2. (a) Prove by induction that a 2 n × 2 n courtyard with a 1 × 1 statue of Bill in a corner can be covered with L-shaped tiles. (Do not assume or reprove the (stronger) result of Theorem 6.1.2 that Bill can be placed anywhere. The point of this problem is to show a different induction hypothesis that works.) Solution. Let P ( n ) be the proposition Bill can be placed in a corner of a 2 n × 2 n courtyard with a proper tiling of the remainder with L-shaped tiles.
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