h1 - 1 Homework I: June 19, 2009 1.9. Find a formula for 1...

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1 Homework I: June 19, 2009 1 . 9 . Find a formula for 1 + n j =1 j ! j ; use induction to prove that your formula is correct. A list of the sums for n = 1 , 2 , 3 , 4 , 5 is 2 , 6 , 24 , 120 , 720. These are factorials; better, they are 2! , 3! , 4! , 5! , 6!. If we write f ( n ) = 1 + n j =1 j ! j , then our guess is f ( n ) = ( n + 1)!. We have already checked the base step n = 1, for f (1) = 2 = 2!. For the inductive step, we must prove f ( n + 1) = 1 + n +1 X j =1 j ! j = ( n + 2)! . Rewrite the middle expression: h n X j =1 j ! j i + ( n + 1)!( n + 1) . The inductive hypothesis says that the bracketed term is ( n + 1)!, and so f ( n + 1) = ( n + 1)! + ( n + 1)!( n + 1) = ( n + 1)!( n + 2) = ( n + 2)! . By induction, f ( n ) = ( n + 1)! for every n 1. 1 . 47 . Given integers a and b (possibly negative) with a 6 = 0, prove that there exist unique integers q and r with b = aq + r and 0 r < | a | . We have already proved this in class when
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h1 - 1 Homework I: June 19, 2009 1.9. Find a formula for 1...

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