SUM OF PRODUCT OF RECIPROCALS OF FIBONACCI NUMBERSA Thesis Submitted in the Partial Fulfillment of...
Transcript of SUM OF PRODUCT OF RECIPROCALS OF FIBONACCI NUMBERSA Thesis Submitted in the Partial Fulfillment of...
SUM OF PRODUCT OF RECIPROCALS OF FIBONACCI NUMBERS
A Thesis Submitted in the Partial Fulfillment of the Requirements of Degree
for
Integrated M.Sc.
In
Mathematics
Submitted by
Kappagantu Prudhavi Nag
Roll Number: 410MA5016
Under the Guidance of
Professor G. K. Panda
Department of Mathematics
National Institute of Technology, Rourkela
May 2015
Page | 2
CERTIFICATE
Dr. Gopal Krishna Panda
Professor of Mathematics May 11, 2015
This is to certify that the project report with title βSUM OF PRODUCT OF RECIPROCALS OF
FIBONACCI NUMBERSβ submitted by Mr. Kappagantu Prudhavi Nag, Roll No. 410MA5016, to
the National Institute of Technology, Rourkela, Odisha for the partial fulfillment of the
requirements of Integrated M.Sc. degree in Mathematics, is a bonafide research work carried
out by him under my supervision and guidance. The content of this report in full or part has not
been submitted to any other Institute or University for the award of any degree or diploma.
Gopal Krishna Panda
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ACKNOWLEDGEMENTS
I would like to express my special appreciation and thanks to my supervisor
Professor G. K. Panda who has been a great mentor for me. I thank him for his suggestion for
providing a beautiful problem in number theory.
I am grateful to Prof. Sunil Kumar Sarangi, Director, National Institute of Technology,
Rourkela for providing excellent facilities in the Institute for carrying out research.
I am thankful to the Head, mathematics and professors of department for their valuable
help during the preparation of this work.
There are no words to explain how grateful I am to my parents for all of their sacrifices
and prayers that they made for me. I also like to thank my brother who helped me in this
research and also gave a few ideas which helped me in successful completion of this project.
I also take this opportunity to thank all of my friends who helped me during the
preparation of this work.
Place: Rourkela
Date:
Kappagantu Prudhavi Nag
Department of Mathematics
National Institute of Technology Rourkela
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ABSTRACT
Fibonacci numbers are the number sequences which follow the linear mathematical
recurrenceπΉ0 = 0, πΉ1 = 1 and πΉπ = πΉπβ1 + πΉπβ2 π β₯ 2. In this work, we study certain sum
formulas involving products of reciprocals of Fibonacci numbers. Sum formulas with
alternating signs are also studied.
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TABLE OF CONTENTS 1. Notations ............................................................................................................................................................6
2. Introduction ......................................................................................................................................................7
i. Mathematics of Fibonacci Numbers .............................................................................................................. 7
ii. Fibonacci Numbers with Negative Indices ................................................................................................ 8
3. Sum of Reciprocals of Fibonacci Numbers with Positive Indices ............................................. 11
i. Order 2 ......................................................................................................................................................................... 11
a. Non-Alternating Sum...................................................................................................................................... 11
b. Alternating Sum ................................................................................................................................................. 13
ii. Sum With Indices in A.P. .................................................................................................................................... 16
a. Non-Alternating Sum of Order 1 ............................................................................................................. 16
b. Alternating Sum of Order 1 ........................................................................................................................ 16
4. Sum of Reciprocals of Fibonacci Numbers with Negative Indices ............................................ 19
i. Order 1 ......................................................................................................................................................................... 19
a. Non-Alternating Sum...................................................................................................................................... 20
b. Alternating Sum ................................................................................................................................................. 20
ii. Order 2 ......................................................................................................................................................................... 20
a. Non-Alternating Sum...................................................................................................................................... 21
b. Alternating Sum ................................................................................................................................................. 21
iii. Sum with Indices in A.P. .................................................................................................................................... 22
a. Non-Alternating Sum of Order 1 ............................................................................................................. 23
b. Alternating Sum of Order 1 ........................................................................................................................ 24
5. References ...................................................................................................................................................... 26
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NOTATIONS The following notations will be frequently used in this thesis.
π½π = β1
πΉπ
ππ=1
πΎπ = β(β1)π
πΉπ
ππ=1
π½π(π) = β1
πΉπ+π
ππ=1
πΎπ(π) = β(β1)π
πΉπ+π
ππ=1
π½πβ = β
1
πΉβπ
ππ=1
πΎπβ = β
(β1)π
πΉβπ
ππ=1
βπ = β1
πΉβπ
ππ=1
ππ = β(β1)π
πΉβπ
ππ=1
βπ(π) = β1
πΉβπβπ
ππ=1
ππ(π) = β1
πΉβπβπ
ππ=1
βπ(0, π) = β1
πΉβππΉβπβπ
ππ=1
ππ(0, π) = β(β1)π
πΉβππΉβπβπ
ππ=1
βπ(π, π) = β1
πΉβπβππΉβπβπ
ππ=1
ππ(π, π) = β(β1)π
πΉβπβππΉβπβπ
ππ=1
βπβ = β
1
πΉββπ
ππ=1
ππβ = β
(β1)π
πΉββπ
ππ=1
βπβ (π) = β
1
πΉββπβπ
ππ=1
ππβ (π) = β
(β1)π
πΉββπβπ
ππ=1
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CHAPTER 1
INTRODUCTION
Fibonacci sequence of numbers is one of the most intriguing number sequence in
mathematics. The series is named after the famous Italian Mathematician Fibonacci of the
Bonacci family. He is also known as the Leonardo of Pisa. The following problem proposed by
Fibonacci himself gave birth to the sequence.
The Fibonacci Problem: Suppose there are two newborn rabbits, one male and one female.
Find the number of rabbits produced in a year if [1]
Each pair takes one month to become mature
Each pair produces a mixed pair every month from the second month
All rabbits are immortal
Solution: For convenience let us assume that the rabbits are born on January 1st and we need
to find the number of rabbits on December 1st. The table below is used to find the solution of
the problem.
From the above table it is evident that the number of rabbits at the end of the year are
144. If observed closely it is observed that the new number is equal to the sum of the previous
two numbers.
MATHEMATICS OF FIBONACCI NUMBERS
The numbers in the bottom row are called the Fibonacci numbers. From the table a
recursive relation is yielded as below
πΉπ = πΉπβ1 + πΉπβ2 , π β₯ 2.
where πΉ0 = 0 and πΉ1 = 1 [2]. Sometimes it is customary to start the Fibonacci numbers from πΉ1
instead of πΉ0. Then the initial two conditions become πΉ1 = 1 and πΉ2 = 1. With any of the above
two conditions the series generated is the same.
No. of Pairs Jan Feb Mar April May June July Aug Sep Oct Nov Dec
Adults 0 1 1 2 3 5 8 13 21 34 55 89
Babies 1 0 1 1 2 3 5 8 13 21 34 34
Total 1 1 2 3 5 8 13 21 34 55 89 144
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It is surprising that Fibonacci numbers can be extracted from Pascalβs triangle. The
above observation was confirmed by Lucas in 1876 when he derived a straightforward formula
to find the Fibonacci numbers [1]
πΉπ = β (π β π β 1
π)
β(πβ1) 2β β
π=0
, π β₯ 1.
In 1843 a French mathematician named Jacques Philippe Marie Binet invented a way to
calculate the ππ‘β Fibonacci numbers. If π =1+β5
2 and π =
1ββ5
2 then πΉπ =
ππβππ
β5 [3].
The above formula shows an interesting aspect that the Fibonacci number can be
written in terms of the golden ratio. Fibonacci numbers appear in many places in both nature
and mathematics. They occur in music, geography, nature, and geometry. They can be found in
the spiral arrangements of seeds of sunflowers, the scale patterns of pine cones, the
arrangement of leaves and the number of petals on the flower.
FIBONACCI NUMBERS WITH NEGATIVE INDICES
The negative subscript of the Fibonacci numbers can be converted to positive subscript
as indicated in [4]. The recurrence relation for the negative Fibonacci numbers is as follows:
πΉβπ = πΉ2βπ β πΉ1βπ, π β₯ 2
The initial conditions for these numbers are πΉ0 = 0 and πΉ1 = 1. It is observed that the
negative Fibonacci numbers have the same initial conditions as of the positive Fibonacci
numbers.
Rabinowitz [5] stated that the alternating general summation of order 1 is given by
πΎπ(π) = β(β1)π
πΉπ+π
π
π=1
= πΎπ+π β πΎπ.
We disprove the above identity using a counter example.
Now let us take π = 3 and π = 5 and find the value of πΎπ(π).
β(β1)π
πΉπ+3
5
π=1
=β1
πΉ4+1
πΉ5+β1
πΉ6+1
πΉ7+β1
πΉ8
πΎ8 =β1
πΉ1+1
πΉ2+β1
πΉ3+1
πΉ4+β1
πΉ5+1
πΉ6+β1
πΉ7+1
πΉ8
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πΎ3 =β1
πΉ1+1
πΉ2+β1
πΉ3
πΎπ+π β πΎπ =1
πΉ4+β1
πΉ5+1
πΉ6+β1
πΉ7+1
πΉ8β πΎπ(π).
It can be checked that the identity for πΎπ(π) is wrong when π is odd and correct when π is even.
Claim: If π > 0 then πΎπ(π) = {πΎπ β πΎπ+π if π is odd,πΎπ+π β πΎπ if π is even.
Proof:
πΎπ+π β πΎπ = β(β1)π
πΉπ
π+π
π=1
ββ(β1)π
πΉπ
π
π=1
= β(β1)π
πΉπ
π+π
π=π+1
.
Case 1. π is odd, π is odd
πΎπ(π) = β(β1)π
πΉπ+π
π
π=1
=β1
πΉ1+π+
1
πΉ2+π+β―+
1
πΉ(πβ1)+π+β1
πΉπ+π.
πΎπ+π β πΎπ = β(β1)π
πΉπ
π+π
π=π+1
=1
πΉ1+π+β1
πΉ2+π+β―+
β1
πΉ(πβ1)+π+
1
πΉπ+π
= β(β1
πΉ1+π+
1
πΉ2+π+β―+
1
πΉ(πβ1)+π+β1
πΉπ+π)
= βπΎπ(π).
Case 2. π is odd, π is even
πΎπ(π) = β(β1)π
πΉπ+π
π
π=1
=β1
πΉ1+π+
1
πΉ2+π+β―+
β1
πΉ(πβ1)+π+
1
πΉπ+π.
πΎπ+π β πΎπ = β(β1)π
πΉπ
π+π
π=π+1
=1
πΉ1+π+β1
πΉ2+π+β―+
1
πΉ(πβ1)+π+β1
πΉπ+π
= β(β1
πΉ1+π+
1
πΉ2+π+β―+
β1
πΉ(πβ1)+π+
1
πΉπ+π)
= βπΎπ(π).
Page | 10
Case 3. π is even , π is odd
πΎπ(π) = β(β1)π
πΉπ+π
π
π=1
=β1
πΉ1+π+
1
πΉ2+π+β―+
1
πΉ(πβ1)+π+β1
πΉπ+π.
πΎπ+π β πΎπ = β(β1)π
πΉπ
π+π
π=π+1
=β1
πΉ1+π+
1
πΉ2+π+β―+
1
πΉ(πβ1)+π+β1
πΉπ+π
= πΎπ (π).
Case 4. π is even , π is even
πΎπ(π) = β(β1)π
πΉπ+π
π
π=1
=β1
πΉ1+π+
1
πΉ2+π+β―+
β1
πΉ(πβ1)+π+
1
πΉπ+π.
πΎπ+π β πΎπ = β(β1)π
πΉπ
π+π
π=π+1
=β1
πΉ1+π+
1
πΉ2+π+β―+
β1
πΉ(πβ1)+π+
1
πΉπ+π
= πΎπ (π).
β
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CHAPTER 2
SUM OF RECIPROCALS OF FIBONACCI NUMBERS WITH
POSITIVE INDICES
The following notations for the alternating and non-alternating sum of ππ‘β order are available
in [5].
π(π1, π2, β¦ , ππβ1, ππ ) = β1
πΉπ+π1πΉπ+π2β¦πΉπ+ππ
π
π=1
π(π1, π2, β¦ , ππβ1, ππ ) = β(β1)π
πΉπ+π1πΉπ+π2β¦πΉπ+ππ
π
π=1
The sum π is called the non-alternating summation of order π. The second sum π is called the
alternating sum of order π. In both the cases 0 < π1 < π2 < β― < ππβ1 < ππ.
ORDER 2
We consider first the problem of finding the following second order sums.
π½π(π, π ) = β1
πΉπ+ππΉπ+π
π
π=1
πΎπ(π, π ) = β(β1)π
πΉπ+ππΉπ+π
π
π=1
NON-ALTERNATING SUM
For π > 0, Rabinowitz [5] got the following formula.
π½π(0, π) = β1
πΉππΉπ+π=
{
1
πΉπβ (
1
πΉπ+2ππΉπ+2π+1β
1
πΉ2ππΉ2π+1)
βπ 2β β
π=1
+πππΉπ if π is odd,
1
πΉπβ(
1
πΉ2πβ1πΉ2πβ
1
πΉπ+2πβ1πΉπ+2π)
π 2β
π=1
if π is even.
π
π=1
where ππ = β1
πΉππΉπ+1
ππ=1 . The above sum is called the non-alternating sum of order 2. The aim is
to find an equivalent expression of π½π(π, π). To do this we use the help of the following result.
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πβπππππ 1: πΉππ 0 < π < π π‘βππ πΉπ (π, π) = πΉπ+π (0, π β π) β πΉπ (0, π β π).
πππππ: We start from the right hand side of the equation and come to the left hand side.
π½π+π(0, π β π) β π½π(0, π β π) = β1
πΉππΉπ+πβπββ
1
πΉππΉπ+πβπ
π
π=1
π+π
π=1
= β1
πΉππΉπ+πβπ
π+π
π=π+1
=1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
1
πΉπ+πβ1πΉπ+πβ1+
1
πΉπ+ππΉπ+π
= β1
πΉπ+ππΉπ+π
π
π=1
= π½π(π, π)
β
By using the above theorem and the formula for π½π(0, π) stated by Rabinowitz [5] the
expression for π½π(π, π) is calculated.
πβπππππ 2: πΌπ 0 < π < π π‘βππ
π½π(π, π) =
{
1
πΉπβπβ (
1
πΉπ+π+2ππΉπ+π+2π+1β
1
πΉπ+2ππΉπ+2π+1)
β(πβπ) 2β β
π=1
+π½π(π, π + 1)
πΉπβπ if (π β π)is odd,
1
πΉπβπβ (
1
πΉπ+2πβ1πΉπ+2πβ
1
πΉπ+π+2πβ1πΉπ+π+2π)
(πβπ) 2β
π=1
if (π β π)is even.
πππππ: We take the help of Theorem 1 to prove this theorem.
π½π+π(0, π β π) =
{
1
πΉπβπβ (
1
πΉπ+π+2ππΉπ+π+2π+1β
1
πΉ2ππΉ2π+1)
β(πβπ) 2β β
π=1
+ππ+ππΉπβπ
if (π β π) is odd,
1
πΉπβπβ (
1
πΉ2πβ1πΉ2πβ
1
πΉπ+π+2πβ1πΉπ+π+2π)
π(πβπ) 2β βπ2β
π=1
if (π β π) is even.
π½π(0, π β π) =
{
1
πΉπβπβ (
1
πΉπ+2ππΉπ+2π+1β
1
πΉ2ππΉ2π+1)
β(πβπ) 2β β
π=1
+πππΉπβπ
if (π β π) is odd,
1
πΉπβπβ (
1
πΉ2πβ1πΉ2πβ
1
πΉπ+2πβ1πΉπ+2π)
(πβπ) 2β
π=1
if (π β π) is even.
Depending upon the parity of (π β π), two different cases are taken into consideration
Page | 13
Case 1. (π β π) π’π¬ π¨ππ
In this case,
π½π(π, π) =1
πΉπβπβ (
1
πΉπ+π+2ππΉπ+π+2π+1β
1
πΉ2ππΉ2π+1β
1
πΉπ+2ππΉπ+2π+1+
1
πΉ2ππΉ2π+1)
β(πβπ) 2β β
π=1
+ππ+π βππ
πΉπβπ
=1
πΉπβπβ (
1
πΉπ+π+2ππΉπ+π+2π+1β
1
πΉπ+2ππΉπ+2π+1) +
ππ+π βππ πΉπβπ
.
β(πβπ) 2β β
π=1
Now,
ππ+π β ππ = β1
πΉππΉπ+1
π+π
π=1
ββ1
πΉππΉπ+1
π
π=1
= β1
πΉππΉπ+1
π+π
π=π+1
=β1
πΉπ+ππΉπ+π+1= π½π(π, π + 1)
π
π=1
π½π(π, π) =1
πΉπβπβ (
1
πΉπ+π+2ππΉπ+π+2π+1β
1
πΉπ+2ππΉπ+2π+1)
β(πβπ) 2β β
π=1
+π½π(π, π + 1)
πΉπβπ .
Case 2. (π β π) is even
π½π(π, π) =1
πΉπβπβ (
1
πΉ2πβ1πΉ2πβ
1
πΉπ+π+2πβ1πΉπ+π+2πβ
1
πΉ2πβ1πΉ2πβ
1
πΉπ+2πβ1πΉπ+2π)
(πβπ) 2β
π=1
=1
πΉπβπβ (
1
πΉπ+2πβ1πΉπ+2πβ
1
πΉπ+π+2πβ1πΉπ+π+2π)
(πβπ) 2β
π=1
.
β
ALTERNATING SUM
Let π > 0 . The following identity is available in from [6].
πΎπ(0, π) = β(β1)π
πΉππΉπ+π=1
πΉπβ(
πΉπβ1
πΉπβπΉπ+πβ1
πΉπ+π)
π
π=1
π
π=1
.
The above is a sum with alternating sign (called an alternating sum) of order 2. We use the
above sum to find a formula for πΎπ(π, π). To achieve this, we use of the following result.
πβπππππ 3: πΌπ 0 < π < π π‘βππ
πΎπ(π, π) = {πΎπ(0, π β π) β πΎπ+π(0, π β π) if π is odd,
πΎπ+π(0, π β π) β πΎπ(0, π β π) if π is even.
πππππ: Observe that
Page | 14
πΎπ+π(0, π β π) β πΎπ(0, π β π) = β(β1)π
πΉππΉπ+πβπ
π+π
π=1
ββ(β1)π
πΉππΉπ+πβπ
π
π=1
= β(β1)π
πΉππΉπ+πβπ.
π+π
π=π+1
We distinguish four cases:
Case 1. π and π are odd
πΎπ(π, π) = β(β1)π
πΉπ+ππΉπ+π=
β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
1
πΉπ+πβ1πΉπ+πβ1
π
π=1
+β1
πΉπ+ππΉπ+π.
πΎπ+π(0, π β π) β πΎπ(0, π β π) = β(β1)π
πΉππΉπ+πβπ
π+π
π=π+1
=1
πΉπ+1πΉπ+2+
β1
πΉπ+2πΉπ+3+β―+
β1
πΉπ+πβ1πΉπ+πβ1+
1
πΉπ+ππΉπ+π
= β(β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
1
πΉπ+πβ1πΉπ+πβ1+
β1
πΉπ+ππΉπ+π)
= βπΎπ(π, π).
Case 2. π is odd and π is even
πΎπ(π, π) = β(β1)π
πΉπ+ππΉπ+π=
β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
β1
πΉπ+πβ1πΉπ+πβ1
π
π=1
+1
πΉπ+ππΉπ+π.
πΎπ+π(0, π β π) β πΎπ(0, π β π) = β(β1)π
πΉππΉπ+πβπ
π+π
π=π+1
=1
πΉπ+1πΉπ+2+
β1
πΉπ+2πΉπ+3+β―+
1
πΉπ+πβ1πΉπ+πβ1+
β1
πΉπ+ππΉπ+π
= β(β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
β1
πΉπ+πβ1πΉπ+πβ1++
1
πΉπ+ππΉπ+π)
= βπΎπ(π, π).
Case 3. π is even and π is odd
πΎπ(π, π) = β(β1)π
πΉπ+ππΉπ+π=
β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
1
πΉπ+πβ1πΉπ+πβ1
π
π=1
+β1
πΉπ+ππΉπ+π.
πΎπ+π(0, π β π) β πΎπ(0, π β π) = β(β1)π
πΉππΉπ+πβπ
π+π
π=π+1
=β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
1
πΉπ+πβ1πΉπ+πβ1++
β1
πΉπ+ππΉπ+π
Page | 15
= πΎπ(π, π).
Case 4. π and π are even
πΎπ(π, π) = β(β1)π
πΉπ+ππΉπ+π=
β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
β1
πΉπ+πβ1πΉπ+πβ1
π
π=1
+1
πΉπ+ππΉπ+π
πΎπ+π(0, π β π) β πΎπ(0, π β π) = β(β1)π
πΉππΉπ+πβπ
π+π
π=π+1
=β1
πΉπ+1πΉπ+1+
1
πΉπ+2πΉπ+2+β―+
β1
πΉπ+πβ1πΉπ+πβ1++
1
πΉπ+ππΉπ+π
= πΎπ(π, π).
β
The following theorem is one of the main results of this chapter.
πβπππππ 4: πΌπ 0 < π < π π‘βππ
πΎπ(π, π) =
{
1
πΉπβπβ(
πΉπ+π+πβ1
πΉπ+π+πβπΉπ+πβ1
πΉπ+π)
πβπ
π=1
if π is odd,
1
πΉπβπβ(
πΉπ+πβ1
πΉπ+πβπΉπ+π+πβ1
πΉπ+π+π)
πβπ
π=1
if π is even.
πππππ: We separate two cases:
Case 1. π is odd
πΎπ(0, π β π) β πΎπ+π(0, π β π) =1
πΉπβπβ(
πΉπβ1πΉπ
βπΉπ+πβ1πΉπ+π
)
πβπ
π=1
β1
πΉπβπβ(
πΉπβ1πΉπ
βπΉπ+π+πβ1πΉπ+π+π
)
πβπ
π=1
=1
πΉπβπβ(
πΉπ+π+πβ1πΉπ+π+π
βπΉπ+πβ1πΉπ+π
)
πβπ
π=1
.
Case 2. π is even
πΎπ+π(0, π β π) β πΎπ(0, π β π) =1
πΉπβπβ(
πΉπβ1πΉπ
βπΉπ+π+πβ1πΉπ+π+π
)
πβπ
π=1
β1
πΉπβπβ(
πΉπβ1πΉπ
βπΉπ+πβ1πΉπ+π
)
πβπ
π=1
=1
πΉπβπβ(
πΉπ+πβ1πΉπ+π
βπΉπ+π+πβ1πΉπ+π+π
) .
πβπ
π=1
β
Page | 16
SUM WITH INDICES IN A.P.
Let β be a natural number. We express certain sums in terms of the following two sums.
π½πβ = β
1
πΉβπ
π
π=1
, πΎπβ =β
(β1)π
πΉβπ
π
π=1
NON-ALTERNATING SUM OF ORDER 1
We consider the problem of finding π½πβ (π) in terms of π½π
β .
πβπππππ 5: πΌπ π > 0 πππ β|π π‘βππ π½πβ (π) = π½
π+π
β
ββ π½π
β
β .
πππππ: Let π =π
β and π = πβ then,
π½π+
πβ
β β π½πβ
β = β1
πΉβπ
π+π
π=1
ββ1
πΉβπ
π
π=1
= β1
πΉβπ
π+π
π=π+1
= 1
πΉβ(π+1)+
1
πΉβ(π+2)+β―+
1
πΉβ(π+πβ1)+
1
πΉβ(π+π)
= 1
πΉβ+π+
1
πΉ2β+π+β―+
1
πΉ(πβ1)β+π+
1
πΉπβ+π
= π½πβ (π).
β
ALTERNATING SUM OF ORDER 1
We consider the problem of finding πΎπβ (π) in terms of πΎπ
β .
πβπππππ 6: πΌπ π > 0 πππ β|π π‘βππ
πΎπβ (π) = {
πΎπβ
β β πΎπ+
πβ
β if π is odd,
πΎπ+
πβ
β βπΎπβ
β if π is even.
Proof: Let π =π
β and π = πβ then,
πΎπ+πβ β πΎπ
β = β(β1)π
πΉβπ
π+π
π=1
ββ(β1)π
πΉβπ
π
π=1
= β(β1)π
πΉβπ
π+π
π=π+1
.
Once again we distinguish four cases.
Case 1. π =π
β and π are odd.
Page | 17
πΎπβ (π) = β
(β1)π
πΉβπ+π
π
π=1
=β1
πΉβ+π+
1
πΉ2β+π+β―+
1
πΉ(πβ1)β+π+
β1
πΉπβ+π.
πΎπ+πβ βπΎπ
β = β(β1)π
πΉβπ
π+π
π=π+1
=1
πΉβ+βπ+
β1
πΉ2β+βπ+β―+
β1
πΉ(πβ1)β+βπ+
1
πΉπβ+βπ
=1
πΉβ+π+
β1
πΉ2β+π+β―+
β1
πΉ(πβ1)β+π+
β1
πΉπβ+π
= β(β1
πΉβ+π+
1
πΉ2β+π+β―+
1
πΉ(πβ1)β+π+
β1
πΉπβ+π)
= βπΎπβ (π).
Case 2. π =π
β is odd, π is even
πΎπβ (π) = β
(β1)π
πΉβπ+π
π
π=1
=β1
πΉβ+π+
1
πΉ2β+π+β―+
β1
πΉ(πβ1)β+π+
1
πΉπβ+π.
πΎπ+πβ βπΎπ
β = β(β1)π
πΉβπ
π+π
π=π+1
=1
πΉβ+βπ+
β1
πΉ2β+βπ+β―+
1
πΉ(πβ1)β+βπ+
β1
πΉπβ+βπ
=1
πΉβ+π+
β1
πΉ2β+π+β―+
1
πΉ(πβ1)β+π+
β1
πΉπβ+π
= β(β1
πΉβ+π+
1
πΉ2β+π+β―+
1
πΉ(πβ1)β+π+
β1
πΉπβ+π)
= βπΎπβ (π)
Case 3. π =π
β is even , π is odd
πΎπβ (π) = β
(β1)π
πΉβπ+π
π
π=1
=β1
πΉβ+π+
1
πΉ2β+π+β―+
1
πΉ(πβ1)β+π+
β1
πΉπβ+π.
πΎπ+πβ β πΎπ
β = β(β1)π
πΉβπ
π+π
π=π+1
=β1
πΉβ+βπ+
1
πΉ2β+βπ+β―+
1
πΉ(πβ1)β+βπ+
β1
πΉπβ+βπ
=1
πΉβ+π+
β1
πΉ2β+π+β―+
1
πΉ(πβ1)β+π+
β1
πΉπβ+π
= πΎπβ (π).
Page | 18
Case 4. π =π
β and π are even
πΎπβ (π) = β
(β1)π
πΉβπ+π
π
π=1
=β1
πΉβ+π+
1
πΉ2β+π+β―+
β1
πΉ(πβ1)β+π+
1
πΉπβ+π.
πΎπ+πβ β πΎπ
β = β(β1)π
πΉβπ
π+π
π=π+1
=β1
πΉβ+βπ+
1
πΉ2β+βπ+β―+
β1
πΉ(πβ1)β+βπ+
1
πΉπβ+βπ
=1
πΉβ+π+
β1
πΉ2β+π+β―+
β1
πΉ(πβ1)β+π+
1
πΉπβ+π
= πΎπβ (π).
β
Page | 19
CHAPTER 3
SUM OF RECIPROCALS OF FIBONACCI NUMBERS WITH
NEGATIVE INDICES
We use the conversion of negative Fibonacci numbers with negative indices to Fibonacci
numbers with positive indices and then derive the identities for these numbers. We use the
following notations.
βπ = β1
πΉβπ
π
π=1
, ππ = β(β1)π
πΉβπ
π
π=1
We first write them in terms of π½π and πΎπ. We use the formula
πΉβπ = (β1)π+1πΉπ
stated in [3]. Then we have
βπ = β1
πΉβπ
π
π=1
= β1
(β1)π+1πΉπ=
π
π=1
β(β1)π+1
πΉπ
π
π=1
= ββ(β1)π
πΉπ
π
π=1
= βπΎπ ,
ππ = β(β1)π
πΉβπ
π
π=1
= β(β1)π
(β1)π+1πΉπ=
π
π=1
β(β1)1
πΉπ
π
π=1
= ββ1
πΉπ
π
π=1
= βπ½π .
β
ORDER 1 We first consider the following sums:
βπ(π) = β1
πΉβπβπ
π
π=1
,
ππ(π) = β(β1)π
πΉβπβπ
π
π=1
.
We write the above in terms of π½π(π) and πΎπ(π).
Page | 20
NON-ALTERNATING SUM
The non-alternating sum of Fibonacci numbers of negative indices of 1st order is,
βπ(π) = β1
πΉβπβπ
π
π=1
.
We write this sum in terms of πΎπ(π) .
βπ(π) = β1
πΉβπβπ
π
π=1
= β1
(β1)π+π+1πΉπ=
π
π=1
β(β1)π+π+1
πΉπ
π
π=1
= (β1)π+1β(β1)π
πΉπ
π
π=1
= (β1)π+1πΎπ(π).
β
ALTERNATING SUM
The alternating sum of Fibonacci numbers of negative indices of 1st order is,
ππ(π) = β(β1)π
πΉβπβπ
π
π=1
.
We express this in terms of π½π(π) .
ππ(π) = β(β1)π
πΉβπβπ
π
π=1
= β(β1)π
(β1)π+π+1πΉπ=
π
π=1
β(β1)π+1
πΉπ
π
π=1
= (β1)π+1β1
πΉπ
π
π=1
= (β1)π+1π½π .
β
ORDER 2 We next consider the following 2nd order sums:
βπ(0, π) = β1
πΉβππΉβπβπ
π
π=1
,
ππ(0, π) = β(β1)π
πΉβππΉβπβπ
π
π=1
,
βπ(π, π) = β1
πΉβπβππΉβπβπ
π
π=1
,
π π(π, π) = β(β1)π
πΉβπβππΉβπβπ
π
π=1
.
We express the above sums in terms of π½π(0, π),πΎπ(0, π), π½π(π, π) and πΎπ(π, π).
Page | 21
NON-ALTERNATING SUM
We consider the following 2nd order non-alternating sums.
βπ(0, π) = β1
πΉβππΉβπβπ
π
π=1
π€βπππ π > 0,
βπ(π, π) = β1
πΉβπβππΉβπβπ
π
π=1
π€βπππ 0 < π < π.
We express the above sums in terms of π½π(0, π) and π½π(π, π) respectively.
First we consider the sum,
βπ(π) = β1
πΉβππΉβπβπ
π
π=1
=β1
(β1)2π+π+2πΉπ=
π
π=1
β(β1)π
πΉππΉπ+π
π
π=1
= (β1)πβ1
πΉππΉπ+π
π
π=1
= (β1)ππ½π(0, π).
β
We next consider,
βπ(π, π) = β1
πΉβπβππΉβπβπ
π
π=1
= β1
(β1)2π+π+π+2πΉπ+ππΉπ+π
π
π=1
= β(β1)π+π
πΉπ+ππΉπ+π
π
π=1
= (β1)π+πβ1
πΉπ+ππΉπ+π
π
π=1
= (β1)π+ππ½π(π, π).
β
ALTERNATING SUM
We consider the following 2nd order alternating sums.
ππ(0, π) = β(β1)π
πΉβππΉβπβπ
π
π=1
π€βπππ π > 0,
ππ(π, π) = β(β1)π
πΉβπβππΉβπβπ
π
π=1
π€βπππ 0 < π < π.
Page | 22
We express the above sums in terms of πΎπ(0, π) and πΎπ(π, π).
First we consider the sum,
ππ(π) = β(β1)π
πΉβππΉβπβπ
π
π=1
=β(β1)π
(β1)2π+π+2πΉπ=
π
π=1
β(β1)π+π
πΉππΉπ+π
π
π=1
= (β1)πβ(β1)π
πΉππΉπ+π
π
π=1
= (β1)ππΎπ(0, π).
β
Next we consider,
ππ(π, π) = β(β1)π
πΉβπβππΉβπβπ
π
π=1
= β(β1)π
(β1)2π+π+π+2πΉπ+ππΉπ+π
π
π=1
= β(β1)π+π+π
πΉπ+ππΉπ+π
π
π=1
= (β1)π+πβ(β1)π
πΉπ+ππΉπ+π
π
π=1
= (β1)π+ππΎπ(π, π).
β
SUM WITH INDICES IN A.P.
Let β be a natural number. We convert the following sums in terms of π½πβ and πΎπ
β .
βπβ =β
1
πΉββπ
π
π=1
, ππβ = β
(β1)π
πΉββπ
π
π=1
.
βπβ = β
1
πΉββπ
π
π=1
= β1
(β1)βπ+1πΉβπ
π
π=1
= β(β1)βπ+1
πΉβπ
π
π=1
= ββ((β1)β)π
πΉβπ
π
π=1
Page | 23
=
{
ββ
(β1)π
πΉβπ
π
π=1
if β is odd,
ββ1
πΉβπ
π
π=1
if β is even.
= {βπΎπ
β if β is odd,
βπ½πβ if β is even.
β
ππβ = β
(β1)π
πΉββπ
π
π=1
= β(β1)π
(β1)βπ+1πΉβπ
π
π=1
= β(β1)βπ+1+π
πΉβπ
π
π=1
= ββ((β1)β+1)π
πΉβπ
π
π=1
=
{
ββ
1
πΉβπ
π
π=1
if β is odd,
ββ(β1)π
πΉβπ
π
π=1
if β is even.
= {βπ½π
β if β is odd,
βπΎπβ if β is even.
β
NON-ALTERNATING SUM OF ORDER 1
We consider the problem of finding 1st order non-alternating sum of Fibonacci number with
negative indices, βπβ (π).
βπβ (π) = β
1
πΉββπβπ
π
π=1
where β > 0 and β|π.
We write the above sum in terms of π½π(π) and πΊπ(π).
Page | 24
βπβ (π) = β
1
πΉββπβπ
π
π=1
=β1
(β1)βπ+π+1πΉβπ+π
π
π=1
=β(β1)βπ+π+1
πΉβπ+π
π
π=1
= (β1)π+1β(β1)βπ
πΉβπ+π
π
π=1
= (β1)π+1β((β1)β)π
πΉβπ+π
π
π=1
= (β1)π+1
{
β
(β1)π
πΉβπ+π
π
π=1
if β is odd,
β1
πΉβπ+π
π
π=1
if β is even.
= {(β1)π+1πΎπ
β (π) if β is odd,
(β1)π+1π½πβ (π) if β is even.
β
ALTERNATING SUM OF ORDER 1
We consider the problem of finding 1st order non-alternating sum of Fibonacci number with
negative indices, ππβ (π).
ππβ (π) = β
(β1)π
πΉββπβπ
π
π=1
where β > 0 and β|π.
We express the above sum in terms of π½π(π) and πΊπ(π).
ππβ (π) = β
(β1)π
πΉββπβπ
π
π=1
= β(β1)π
(β1)βπ+π+1πΉβπ+π
π
π=1
= β(β1)βπ+π+1
πΉβπ+π
π
π=1
Page | 25
= (β1)π+1β(β1)βπ+π
πΉβπ+π
π
π=1
= (β1)π+1β((β1)(β+1))
π
πΉβπ+π
π
π=1
= (β1)π+1
{
β
1
πΉβπ+π
π
π=1
if β is odd,
β(β1)π
πΉβπ+π
π
π=1
if β is even.
= {(β1)π+1π½π
β (π) if β is odd,
(β1)π+1πΎπβ (π) if β is even.
β
Page | 26
REFERENCES
[1] T. Koshy, Elementary Number Theory with Application, London: Academic Press, 2007.
[2] M. Krzywkowski, "New Proofs of Some Fibonacci Identities," International Mathematical Forum, vol. 5, no. 18, pp. 869 - 874, 2010.
[3] D. Kalman and R. Mena, "The Fibonacci Numbers-Exposed," Mathematics Magazine, vol. 76, no. 3, pp. 167-181, 2003.
[4] S. Rabinowitz, "Algorithmic Manipulation of Fibonacci Identities," Kluwer Academic Publishers, vol. 6, pp. 389-408, 1996.
[5] S. Rabinowitz, "Algorithmic Summation of Reciprocals of Products of Fibonacci Numbers," The Fibonacci Quarterly, vol. 37, pp. 122-127, 1999.
[6] I. J. Good, "A Symmetry Property of Alternating Sums of Products of Reciprocals," The Fibonacci Quarterly, vol. 32, no. 3, pp. 284-287, 1994.
[7] L. Carlitz, "Reduction Formulas for Fibonacci Summations," The Fibonacci Quaterly, vol. 9, no. 5, pp. 510-511, 1971.
[8] D.Burton, Elementary Number Thoery, New York: MacGraw Hill Higher Education, 2007.
[9] M. Jastrzebska and A. Grabowski, "Some Properties of Fibonacci Numbers," Formalized Mathematics, vol. 12, no. 3, pp. 307-312, 2004.
[10] A. T. Benjamin and J. J. Quinn, "Recounting Fibonacci and Lucas Identities," College Math, vol. 30, no. 5, pp. 167-181, 2003.