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)?
In this series except for 1st and 2nd term
T (n) = T (n-1)/10 + T(n-2)/100
Therefore
T (1) = 0.1
T (2) = 0.01
T (3) = T (2)/10 + T (1)/100
T (4) = T (3)/10 + T (2)/100
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T (n) = T (n-1)/10 + T(n-2)/100
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If we add all the above terms (take 0.01 = T(1)/10 then
Sum = 0.1 + Sum/100 + Sum/10
implies Sum = 10/89 = 0.11235955
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Posted by Sachin
on 2005-03-09 21:46:30 |
Solution with explaination
