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Unformatted text preview: Graded problems: 1.4 (15 points), 2.3 (10 points), 3.6 (10 points), 4.8 (15 points); the other problems were marked for completion only and worth five points each (except that 1.12 and 4.1, which were somewhat longer, were worth ten points). 1.3. To proceed by induction, we prove the result for n = 1 and then, assuming the result for n , derive the result for n + 1. n = 1 : 1 3 = 1 = 1 2 . n implies n + 1: Suppose the result holds for a given natural n , i.e. 1 3 + 2 3 + . . . + n 3 = (1 + 2 + . . . + n ) 2 . Then 1 3 + 2 3 + . . . + n 3 + ( n + 1) 3 = (1 + 2 + . . . + n ) 2 + ( n + 1) 3 = (1 + 2 + . . . + n ) 2 + [( n + 1) 3 − ( n + 1) 2 ] + ( n + 1) 2 = (1 + 2 + . . . + n ) 2 + ( n + 1) 2 [ n + 1 − 1] + ( n + 1) 2 = (1 + 2 + . . . + n ) 2 + 2( n + 1) n ( n + 1) / 2 + ( n + 1) 2 = (1 + 2 + . . . + n ) 2 + 2( n + 1)(1 + 2 + . . . + n ) + ( n + 1) 2 (by Example 1) = (1 + 2 + . . . + ( n + 1)) 2 (by the rule ( a + b ) 2 = a 2 + 2 ab + b 2 , where a = 1 + 2 + . . . + n and b = n + 1) showing the result for n + 1, so induction gives us exactly what we need. 1.4(a): n = 1: The expression is simply 1. n = 2: The expression is 1 + 3 = 4. n = 3: The expression is 1 + 3 + 5 = 9. n = 4: The expression is 1 + 3 + 5 + 7 = 16. For each of these, we note 1+3+ . . . +(2 n − 1) = n 2 , so this is our proposed formula. (b): We already have the n = 1 result from part (a); therefore, it suffices to show the formula for n +1 given the formula for some n . As 2( n +1) − 1 = 2 n − 1, we have (assuming the result for n ) 1 + 3 + . . . + (2( n + 1) − 1) = 1 + 3 + . . . + (2 n − 1) + (2 n + 1) = n 2 + 2 n + 1 = ( n + 1) 2 as desired, giving us the formula. 1 1.10. We could prove this formula by induction; however, there is a quicker way. We observe that (2 n +1)+(2 n +3)+ . . . +(4 n − 1) = (1+3+ . . . +(4 n − 1)) − (1+3+ . . . +(2 n − 1)) = (1 + 3 + . . . + 2(2 n ) − 1) − (1 + 3 + . . . + (2 n − 1)) = (2 n ) 2 − n 2 (by Problem 1.4’s formula) = 4 n 2 − n 2 = 3 n 2 as desired. 1.12(a). n = 1: The left hand side is ( a + b ) 1 = a + b ; the right hand side is a 1 + b 1 = a + b so the theorem holds. n = 2: The left hand side is ( a + b ) 2 = a 2 + 2 ab + b 2 ; the right hand side is a 2 + 2 ab + b 2 so the theorem holds. n = 3: The left hand side is ( a + b ) 3 = ( a 2 + 2 ab + b 2 )( a + b ) = a 3 + 2 ab 2 + ab 2 + a 2 b + 2 ab 2 + b 3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 ; the right hand side is a 3 + 3 a 2 b + 1 / 2 ∗ 3 ∗ 2 ∗ ab 2 + b 3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 so the theorem holds....
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 Spring '09
 WEISBART
 Supremum, Order theory, lower bound, upper bound, nCk

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