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Unformatted text preview: Math 55  Homework 3 Solutions Wednesday, July 6, 2011 1. From section 4.1, do problem 70. (a) At step n + 1 , we would get: 1 2 3 4 ... 2 n 1 2 n 2 n + 1 2 n + 2 < 1 √ 3 n · 1 1 + 1 2 n +1 = 1 q 3 n + 12 n 2 +9 n 4 n 2 +4 n +1 6≤ 1 √ 3 n + 3 Since 12 n 2 +9 n 4 n 2 +4 n +1 < 3 . (b) For n = 1 , we verify that 1 2 < 1 √ 3 . For n ≥ 2 , we will prove by induction that 1 2 3 4 ... 2 n 1 2 n < 1 √ 3 n + 1 For the base case of n = 2 , Note that ( 8 3 ) 2 = 64 9 > 7 , so: 1 2 3 4 = 3 8 < 1 √ 7 = 1 p 3(2) + 1 Now assume the inequality holds for n , and apply it to the n + 1 case: 1 2 3 4 ... 2 n 1 2 n 2 n +1 2 n +2 < 1 √ 3 n +1 · 1 1+ 1 2 n +1 = 1 r 3 n +1+ 12 n 2 +13 n +3 4 n 2 +4 n +1 < 1 r 3 n +1+ 12 n 2 +12 n +3 4 n 2 +4 n +1 = 1 √ 3( n +1)+1 Which completes the induction, and gives the originally desired result. 2. From section 4.1, do problem 76. First, notice it is possible to tile 5 × 5 if the hole is in the corner: Base case 1: n = 7 : By rotating and reflecting, we can put the hole in one of the three 2 × 2 boxes, A, B, or C: And we can finish that 2 × 2 block with one more triomino. Base case 2: n = 11 : Draw a 7 × 7 box in a corner that contains the hole. This is possible since 7 ≥ 11 2 . Draw a 5 × 5 box in the opposite corner. These two boxes overlap in a single square. The remaining two corners contain 4 × 6 boxes. Cover the 7 × 7 missing the hole using base case 1. Cover the 5 × 5 box without it’s overlapping corner as above. The two 4 × 6 boxes can be covered with 2 × 3 boxes made out of two triominoes each. Inductive step: Let n be odd, not divisible by 3, and greater than 11. Assume that if 5 < k < n , and k is odd and not divisible by 3, then any k × k square with a 1 × 1 square removed can be tiled with triominoes....
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 Summer '08
 STRAIN
 Math, Recurrence relation, Fibonacci number, Golden ratio

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