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Fibonacci Fractions (Posted on 2005-03-09) Difficulty: 3 of 5
What is the sum of 0.1+ 0.01+ 0.002+ 0.0003+ 0.00005+ 0.000008+ ..., where each term is the n-th Fibonacci number, shifted n places to the right (that is, divided by 10^n)?

See The Solution Submitted by e.g.    
Rating: 3.0000 (2 votes)

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Solution Puzzle Solution: Method III Comment 17 of 17 |
(In reply to Puzzle Solution: Method II by K Sengupta)

Let S = 0.1+ 0.01+ 0.002+ 0.0003+ 0.00005+ 0.000008+ ...

Then,
(10^4)*S  =  1123 +  0.5 + 0.08 + 0.013 + 0.0021 + 0.00034+ ...

30*S =             3 +  0.3 + 0.06 + 0.009 + 0.0015 + 0.00024 + ...

-> 9970*S = 1120 + 0.2 + 0.02 + 0.004 + 0.0006 + 0.00010+ ...
= 1120 + 2(0.1+ 0.01+ 0.002+ 0.0003+ 0.00005+..... )
= 1120 + 2*S

or, (9970 - 2)*S = 1120

or, S = 1120/9968 = 10/89


  Posted by K Sengupta on 2008-05-01 06:32:52
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