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ma017-2010-sol4

# ma017-2010-sol4 - MA017 2010 HOMEWORK 4(1 Show that n...

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MA017 2010 HOMEWORK 4 (1) Show that n X k =0 ( - 1) k ( n k ) ( k + 1) 2 = 1 n + 1 n +1 X k =1 1 k . Consider (1 - x ) n = ( - 1) k ( n k ) . Integrate both sides, then divide both sides by x and integrate again (be careful with the boundary terms). (2) Prove that all terms of the sequence a 1 = a 2 = a 3 = 1 , a n +1 = 1 + a n - 1 a n a n - 2 are integers. This was a Putnam problem, but I don’t remember the year. You can solve it by looking for a more symmetric form recursion, which leads naturally to the discovery of an invariant, as follows. Looking at the given recursion for n and n + 1: a n +1 a n - 2 = 1 + a n - 1 a n , 1 + a n a n +1 = a n +2 a n - 1 . Adding these two equations, we get a n +1 ( a n - 2 + a n ) = a n - 1 ( a n + a n +2 ) , i.e. that a n - 2 + a n a n - 1 = a n + a n +2 a n +1 . Thus b n := a n + a n +2 a n +1 satisﬁes the recursion b n +2 = b n . We have a 1 = 1 ,a 2 = 1 ,a 3 = 1 ,a 4 = 2, so that b 1 = 2 and b 2 = 3; thus in general b 2 n = 1 ,b 2 n +1 = 3. The recursion

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ma017-2010-sol4 - MA017 2010 HOMEWORK 4(1 Show that n...

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