100/89= 1.1235..., which includes the first five Fibonacci numbers.

10000/9899=1.010203050813213455... includes the first ten Fibonacci numbers.

1000000/998999=1.001002003005... produces the first 15 Fibonacci numbers.

If you add two zeros in the numerator and two nines (one at the beginning, one at the end) in the denominator, does this Fibonacci production go on?

The rule that defines the Fibonacci numbers is that a(n)=a(n-1)+a(n-2) -- so we want to prove a version of that for numbers of the form F(k)=10^2k/(10^2k-10^k-1) (the numbers in the problem)

The fibonacci numbers are k digits long, so multiplying by 10^k would shift each number over one space. Then we just need to get rid of the leading ones. (ie 1.1235+10*(1.1235-1)=(1.1235-1.1)*100) or (1.1235..+1.235..=2.35..) It can be seen that just the first 1 needs to be subtracted, since the other already exists on the left side and right side.

Thus, F(k)+10^k*(F(k)-1)=10^(2k)*(F(k)-1-(10^-k))

F(k)+10^k*F(k)-10^k=10^(2k)*F(k)-10^(2k)-10^(k)
To find F(k), put all F(k) terms on one side and constant terms on the other side: 10^(2k)=(10^2k-10^k-1)*F(k) and divide to get F(k)=10^(2k)/(10^2k-10^k-1) which is just what we wanted.

Posted by Gamer on 2007-02-22 18:11:01