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Unformatted text preview: full. After the first part of the first step, the bottom cup will have 1  (1/k) = (k1)/k cups of water remaining. The addition of the wine from the top cup in the secod part of the first step will not change this amount. Each succesive cycle of the two steps will result in the amount of water remaining being diminished by 1/k of what was there, so the amount remaining is, again, 1  (1/k) of the previous amount. This means that after n steps, you have [ (k1)/k ] n cups of water remaining in the bottom cup. Specifically, after all k steps are done, youll have [IMAGE] cups remaining in the bottom cup. To complete the problem, take k to be very large. The value above will approach the value of the limit, which can be computed exactly using, for example, LHpitals Rule. This value is 1/e....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.
 Spring '10
 AlbertoDelgado
 Combinatorics

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