As we have seen when we add up the terms of the Fibonacci sequence in the following decimal point form.01 + .001 + .0002 + .00003 + .000005 + .0000008 + .00000013 + .000000021 +
the sum = 1/89.
Equally fascinating results apply for the corresponding series where we combine more than two terms to generate the next term in the sequence.
Thus in the 3-term case, the relevant series is 1, 1, 2, 4, 7, 13, 24, 44, 81, 149,
Therefore the decimal point series is
.01 + .001 + .0002 + .00004 + .000007 + .0000013 + .00000024 + .000000044 + .0000000081 +
The sum of this series = 1/88.9
In the one term case where the series is 1, 1, 1, 1, 1, 1,
The resulting point series = .01 + .001 + .0001 + .00001 + .000001 +
= .01111111.. = 1/90
90 = 100 - 10
In the two term case, where the result = 1/89,
89 = 100 - 10 - 1.
Now we see that in the three-term case the result =1/ 88.9,
Here 88.9 = 100 - 10 - 1 - .1.
Thus suggests that the sum of the decimal point series of terms in the 4-term case is 1/88.89
Where 88.89 = 100 - 10 - 1 - .1 - 01.
The four term-series is 1, 1, 2, 4, 8, 15, 29, 56, 108, 208,
Therefore the relevant point series is
.01 + .001 + .0002 + .00004 + .000008 + .0000015 + .00000029 + .000000056 +
= 1/88.89
The number of 8's is one less than the number of terms combined in series.
Therefore for example in the seven-term case, the sum of decimal point terms
= 88.88889
The series is 1, 1, 2, 4, 8, 16, 32, 64, 127, 253, 504, 1004, 2000, 3984, 7936,..
Therefore the relevant point series is
.01 + .001 + .0002 + .00004 + .000008 + .0000016 + .00000032 + .000000064 + .0000000127 + .00000000253 + .000000000504 + .0000000001004 + .00000000002000 +
= 1/88.88889
Once again these results are fully transferable across number bases.
Thus in base 8 for example the 3-term series is
1, 1, 2, 4, 7, 15, 30, 54, 101, .
Therefore the sum of the relevant point terms = reciprocal of 100 - 10 - 1 - .1 (in base 8)
= 1/66.7
The point term series is
.01 + .001 + .0002 + .00004 + .000007 + .0000015 + .00000030 + .000000054 + ..
= 1/66.7
We can also generate similar type results to the two-term case when we alternately add and subtract terms in the point series.
In the two-term case we saw that
.01 - 001 + .0002 - .00003 + .000005 - .0000008 + .00000013 - .000000021 +
= 1/109
Again in the one-term case where the series is 1, 1, 1, 1, 1, 1,
The resulting point series = .01 - .001 + .0001 - .00001 + .000001 -
= .01111111.. = 1/110
110 = 100 + 10
In the two-term case, where the result = 1/109,
109 = 100 + 10 - 1.
In the three-term case the sum of alternating point terms is
.01 - .001 + .0002 - .00004 + .000007 - .0000013 + .00000024 - .000000044 + .0000000081 -
= 1/109.1 where 109.1= 100 + 10 - 1 + .1
What is involved here is a process by which the sign for each term switches - after the first two terms - in an ordered fashion. Therefore by extension the point sequence associated with the four-term case = the reciprocal of 100 + 10 - 1 + .1 - .01 = 109.09
Thus .01 - .001 + .0002 - .00004 + .000008 - .0000015 + .00000029 - .000000056 +
= 109.09
The seven-term case here is
.01 - .001 + .0002 - .00004 + .000008 - .0000016 + .00000032 - .000000064 + .0000000127 - .00000000253 + .000000000504 - .0000000001004 + .00000000002000 -
= 1/109.09091
With n terms the number of terms in formula = n + 1.
Therefore the sum of terms = the reciprocal of 100 + 10 - 1 + .1 - .01 + .001 - .0001 + .00001
= 1/109.09091
Once again these results can be extended across number bases. In the next article we will look at the binary case.
When we divide 109.09091 by the corresponding number in the positive series 88.88889, the result 1.227272722 displays remarkable palindromic qualities. This is not accidental as the palindromic features improve with the numbers associated with the sums of the higher order term sequences.
The reciprocal .81481481.. also displays similar palindromic tendencies which likewise improves as n increases.
1.227272722.. - 1 = .227272722 and the reciprocal of this = 4.4