All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Puzzle Solution: Method II | Comment 16 of 17 |
(In reply to Puzzle Solution: Method I by K Sengupta)

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

Then,
1000*S = 112 + 0.3 + 0.05 + 0.008 + 0.0013 + ......
20*S =     2 + 0.2 + 0.03 + 0.005 + 0.0008 + ......

or, (1000*S - 20*S) = 110 + (0.1+ 0.01+ 0.002+ 0.0003+ 0.00005+.....)
or, 980*S = 110 + S
or, 979*S = 110
or, S = 110/979 = 10/89

Edited on May 1, 2008, 6:31 am
  Posted by K Sengupta on 2008-05-01 06:31:10

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information