Once again the Fibonacci Series is

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1589, ....

As we have seen there is a well-ordered series of fascinating connections between phi and the respective terms of the Lucas series1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 621, 943, ….

Thus

Phi - 1/Phi = 1 (the first term of Lucas Series)

Phi

^{2}+ 1/Phi^{2}= 3 (the second term of Lucas series)Phi

^{3}- 1/Phi^{3 }= 4 (the third term of Lucas Series)Phi

^{4}+ 1/Phi^{4}= 7 (the fourth term of Lucas Series)Phi

^{5}- 1/Phi^{5}= 11 (the fifth term of Lucas Series)So in general format

Phi

^{n}(+ or -) 1/Phi^{n }= T_{n }(where Tn represents the nth term of the Lucas Series).When n is even we add the respective values of Phi.

When n is odd, we subtract the respective values of Phi.

Thus for example T

_{11}= Phi^{11}- 1/Phi^{11 }= 199 (which is the 11^{th}term in the series).

Now here is a fascinating relationship with the above which can be stated as follows

If we take the sum of any sequence of n consecutive terms of the Fibonacci Series and then divide this total by the sum of the n terms directly preceding this sequence, the corresponding result approximates Phi

^{n}.

Once again when this result is combined with its resiprocal in the manner described above we obtain the corresponding nth term in the related Lucas Series.Thus when n = 1, let us for example take the 15th term of the Fibonacci is 610. When we divide this by the preceding term we get 610/377 = 1.618037.. which approxmates Phi

^{1 }(n = 1).When we subtract the reciprocal (377/610 = .618032...) the result approximates to 1 (i.e. T

_{1 }in the Lucas Series).When n = 3 for example the sum of three consecutive terms (e.g. 15th, 16th and 17th terms) = 610 + 987 + 1597 = 3194. the sum of the three preceding terms (i.e. 12th, 13th and 14th) = 754.

The ratio of these two sums = 3194/754 =4.236074.... (which approximates Phi

^{3}).Once again when we subtract the reciprocal (754/3194 = .236067...) the result approximates 4 (i.e. T

_{3 }in the Lucas Series).

(The accuracy of such approximations improves with the use of higher terms in theFibonacci Series).Finally to illustrate let n = 6. We will take to illustrate the sum of the 12th, 13th, 14th, 15th, 16th and 17th terms i.e. 144 + 233 + 377 + 610 + 987 + 1597 = 3948). The sum of the 6 preceding terms (6th, 7th, 8th, 9th, 10th and 11th = 8 + 13 + 21 + 34 + 55 + 89 = 220).

The ratio of the sum of these consecutive sequences = 3948/220 = 17.94545.... This result approximates Phi^{6}.When we add the reciprocal in this case (i.e. 220/3948 = .055728...) as n is even, the result approximates to 18 (i.e. T

_{6 }in the Lucas Series).Alternatively since T

_{n - 1 }+ T_{n + 1 }of the Fibonacci = T_{n }in the Lucas Series, we can express all these results in terms of the original Fibonacci Series.So if we take the sum of any sequence of n consecutive terms of the Fibonacci Series, divide this total by the sum of the n terms directly preceding this sequence and then alternatively add or subtract its reciprocal (depending on whether n is even or odd respectively), the result approximates to

T_{n - 1}+ T_{n + 1 }(of the same Fibonacci Series).

Surprisingly if we likewise take the sum of n consecutive terms, where the signs of terms alternate in an orderly alternate positive and negative fashion, and then divide by the sum of the n preceding terms (where again signs of terms alternate), and again ass or subtract its reciprocal (in the manner already desdcribed) the same result approximates (i.e. T

_{n }of the related Lucas, or T_{n - 1 }T + T_{n + 1 }of the same Fibonacci Series.

To illustrate let us take the case above where n = 6, this time combining successive terms in alternate positive and negative fashion.

So combining the 12th, 13th, 14th, 15th, 16th and 17th terms in alternate positive and negative fashion we get 144 - 233 + 377 - 610 + 987 - 1597 = -932). The combination of the 6 preceding terms (6th, 7th, 8th, 9th, 10th and 11th = 8 - 13 + 21 - 34 + 55 - 89 = -52).

The ratio of the sum of these consecutive sequences = -932/-52 = 17.923.... This result approximates Phi^{6}.^{ }When^{ }we add 1/Phi^{6 }the result approximates 18 (i.e T_{6 }of the related Lucas, or T_{5 }+ T_{ 7 }of the same Fibonacci Series)^{ }.

Amazingly, if we take the sum of any sequence of the n consecutive reciprocal terms of the Fibonacci Series, divide this total by the corresponding sum of the the n preceding reciprocal terms, and procede in the same manner we obtain the reciprocal of the previous results (i.e 1/T

_{n }of the related Lucas,

or 1/T_{n - 1 }+ 1/T_{ n + 1 }of the same Fibonacci Series).For example using the same example as above (i.e. where n = 6), adding the reciprocals of the 12th, 13th, 14th, 15th, 16th and 17th terms gives 1/144 + 1/233 + 1/377 + 1/610 + 1/987 + 1/1597 = .017167499...

The corresponding sum of the 6 preceding terms (6th, 7th, 8th, 9th, 10th and 11th) = 1/8 + 1/13 + 1/21 + 1/34 + 1/55 + 1/89 = .308371662.

Dividing the two results we get .05567145.. which is the reciprocal of 17.962...

Once again this is an approximation of Phi

^{6 }which when combined with^{ }1/Phi^{6}approximates to 18

(i.e T_{6 }of the related Lucas, or T_{5 }+ T_{ 7 }of the same Fibonacci Series).A remarkable consequence of this result is that

(1/T_{k }+ 1/T_{k + 1}... + 1/T_{k + n - 1 })/(1/T_{k -1 }+ 1/T_{k - 2}... + 1/T_{k - n}) approximates to1/(T

_{k }+ T_{k + 1}... + T_{k + n - 1}) divided by 1/(T_{k - 1}+ T_{k - 2 }... + T_{k - n})Thus if for example n = 3, and we take the reciprocals of three consecutive terms of the Fibonacci

e.g. 1/89 + 1/144 + 1/233 and then divided this by the sum of reciprocals of three preceding terms i.e.

1/21 + 1/34 + 1/55 we obtain .02247724499./.0952126305. which approximates to 1/Phi^{3}.Now 1/(89 + 144 + 233) divided by 1/(21 + 34 + 55) = .0021459227./.009090909 which again approximates to 1/Phi

^{3}.