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

 Fibonacci Fractions (Posted on 2005-03-09)
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.)
 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
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 (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information