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-termcase, the relevant series is1, 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

1/88.89Where

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

=

88.88889The 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 8for examplethe 3-termseries 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)=

1/66.7The 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-termcase 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-termcase where the series is1, 1, 1, 1, 1, 1,The resulting point series =

.01 - .001 + .0001 - .00001 + .000001 -=

.01111111..=1/110

110 = 100 + 10

In the

two-termcase, where the result =1/109,

109 = 100 + 10 - 1.

In the

three-termcase the sum of alternating point terms is

.01 - .001 + .0002 - .00004 + .000007 - .0000013 + .00000024 - .000000044 + .0000000081 -

= 1/109.1where109.1= 100 + 10 - 1 + .1What 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-termcase = the reciprocal of100 + 10 - 1 + .1 - .01 = 109.09Thus

.01 - .001 + .0002 - .00004 + .000008 - .0000015 + .00000029 - .000000056 +=

109.09

The

seven-termcase here is

.01 - .001 + .0002 - .00004 + .000008 - .0000016 + .00000032 - .000000064 + .0000000127 - .00000000253 + .000000000504 - .0000000001004 + .00000000002000 -

= 1/109.09091With

n termsthe number of terms in formula= n + 1.Therefore

the sum of terms =the reciprocal of100 + 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

109.09091by the corresponding number in the positive series88.88889,theresult 1.227272722displays remarkablepalindromicqualities. 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 = .227272722and the reciprocal of this =4.4