mdm12Section4_4_OddSolutionsFinal

# mdm12Section4_4_OddSolutionsFinal - Q1-3 Q7 Q13-15 Q5 Q9-11...

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Q1-3 Q5 Q7 Q9-11 Q13-15 Q17-19 CHAPTER 4 Section 4.4, pp. 251–253 Solutions to Odd Number Problems 1. 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 1 9 36 84 126 126 84 36 9 1 1 10 45 120 210 252 210 120 45 10 1 1 11 55 165 330 462 462 330 165 55 11 1 1 12 66 220 495 792 924 792 495 220 66 12 1 3. a) sum(row 12) = 2 12 = 4096 b) sum(row 20) = 2 20 = 1 048 576 c) sum(row 25) = 2 25 = 33 554 432 d) sum(row n 1) = 2 n 1 Top

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5. a) Row Sum/Difference Result 0 1 1 1 1 1 0 2 1 – 2 + 1 0 3 1 – 3 + 3 – 1 0 4 1 – 4 + 6 – 4 + 1 0 5 1 – 5 + 10 – 10 + 5 – 1 0 6 1 – 6 + 15 – 20 + 15 – 6 + 1 0 b) 0 Row Sum/Difference Result 6 1 – 6 + 15 – 20 + 15 – 6 + 1 0 7 1 – 7 + 21 – 35 + 35 – 21 + 7 – 1 0 8 1 – 8 + 28 – 56 + 70 – 56 + 28 – 8 + 1 0 c) 0 Top
7. a) 11 1 = 11 11 2 = 121 11 3 = 1331 11 4 = 14 641 The digits of the first four powers of 11 match the numbers in the rows of Pascal's triangle. b) 11 5 = 161 051 The fifth row of Pascal's triangle is 1 5 10 10 5 1. You could show row 5 as a power of 11 by expressing 6 as 5 + 1 and the following 1 as 0 + 1.

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## This note was uploaded on 01/23/2012 for the course MATHS MDM-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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mdm12Section4_4_OddSolutionsFinal - Q1-3 Q7 Q13-15 Q5 Q9-11...

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