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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 Explained solution | Comment 10 of 17 |
There is a famous formula that gives the n-th Fibonacci number:  (((1+√5))^n -((1-√5))^n)/√5.

We are then asked for the sum, for n=0 to infinity, of f(n)/10^n.

This equals the sum of two geometric series,  so applying the usual formula we get (1/(1-(1+√5)/20) - 1/(1-(1+√5)/20))/√5, which works out to be 10/89.

Edited on March 9, 2005, 9:19 pm
  Posted by Federico Kereki on 2005-03-09 21:12:57

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