Unformatted text preview: an+l = an + anI, we obtain a~~l = 1 + a~:l As n + 00, the left side has limit k while the right side has limit 1 + i. The solutions to k = 1 + i are the solutions to k 2 k 1 = 0, namely, k = 1± 2 Y5. Since k > 0, k = 1+ 2 Y5 is the golden mean. 00 fk 30. Let S = L k . Then 1 100 ThusS 1=( 1 1) 1 100 10000 100 ' 9899 S = _1_ d S = 100 = .01010203050813213456. 10000 100 an 9899 The first ten terms of the Fibonacci sequence are clearly visible. Exercises 6.1 1. [BB] Let B be the set of connoisseurs of Canadian bacon, and A the set of those who like anchovies. (a) This asks for IB U AI = IBI + IAI  IB n AI = 10 + 7 6 = 11. (b) This asks for IB " AI = IBI  IB n AI = 10 6 = 4....
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 Summer '10
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 Graph Theory, Trigraph, Recurrence relation, Fibonacci number, IB n AI, recurrence relation an+l

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