Fibonacci
Numbers
By: Sara Miller
Advisor: Dr. Mihai Caragiu
Abstract
v We will investigate various ways of
proving identities involving Fibonacci
Numbers, such as, induction, linear
algebra (matrices), and combinatorics
(0-1 sequences).
vWe will also look at one educational
activity.
Summary
v
v
v
v
v
1.
2.
3.
4.
5.
Introduction
Mathematical Induction
Linear Algebra (matrices)
Combinatorics (0-1 sequences)
Educational Activity
1.
Introduction
v Leonardo Pisano
O
O
O
O
Born around 1175 in Pisa, Italy
His nickname was Fibonacci
He traveled extensively with his father
Wrote book: Liber Abbaci in 1220
• Presented a numerical series now referred
to as the Fibonacci numbers
Fibonacci Numbers
v
1, 1, 2 , 3, 5, 8, 13, 21, 34, 55,
89, 144, 233, …
Rabbit Problem
v “A certain man put a pair of rabbits in a
place surrounded by a wall. How many
pairs of rabbits can be produced from
that pair in a year if it is supposed that
every month each pair begets a new pair
from which the second month on becomes
productive?” (Liber Abbaci, chapter 12, p.
283-4)
Chart of Rabbit Problem
Month
Adult Pairs Baby Pairs Total Pairs
January
1
0
1
February
1
1
2
March
2
1
3
April
3
2
5
May
5
3
8
June
8
5
13
July
13
8
21
August
21
13
34
September
34
21
55
October
55
34
89
November
89
55
144
December
144
89
233
January
233
144
377
The Fibonacci Linear
Recurrence
Initial Conditions:
f = 1 and f = 1
1
2
Recurrence relation:
fn = f
n−1
+f
n−2
n
1
fn 1
2
1
3
2
4
3
5
5
6
8
7
13
2. Mathematical Induction
2.1
Sum of Squares of Fibonacci Numbers:
f 2 + f 2 + ... + f n2 = f n f
n+1
1
2
, n ≥1
• I will initially carry out the proof of this identity
by induction. Then I will provide a “visual” proof.
Proof by induction:
(a) For n =1 the formula takes the form
f 2= f f
1 1 2
⎡
⎢
⎢⎣
⎤
12 = (1)(1) thus 1=1⎥
⎥⎦
Thus the first part of the induction
(basis step) is finished.
(b) Assuming the formula holds true for n, we will
prove it for n+1.
Therefore,
f 2 + f 2 +...+ f n2 = f n f
1
2
n+1
And we will prove
f 2 + f 2 + ...+ f n2 + f 2 = f
f
1
2
n+1 n+1 n+2
Indeed,
f 2 + f 2 + ...+ f n2 + f 2
1
2
n+1
By inductive hypothesis
=f f
+f2
n n +1 n+1
=f
=f
⎛
⎜
⎜
⎝
n+1
f
fn + f
n+1 n+2
n+1
⎞
⎟
⎟
⎠
By f
n+1
+ fn = f
n+2
Thus the induction is complete, and I have proved that
the formula holds for all n.
QED.
Visual Proof
f 2 + f 2 + ...+ f n2 = f n f
1
2
n+1
2.2 Sum of Odd Fibonacci Numbers:
=f
f + f + f + ...+ f
1 3 5
2n−1 2n
, n ≥1
2.3 Sum of Even Fibonacci Numbers:
f + f + f + ...+ f = f
−1 , n ≥ 0
0 2 4
2n 2n+1
2.4
Sum of Products of Consecutive Fibonacci
Numbers:
f f + f f + f f + ...+ f
f = f 2 , n ≥1
1 2 2 3 3 4
2n−1 2n 2n
Proof:
(a) For n =1 the formula takes the form
f f = f2
1 2 2
⎡ ⎛ ⎞⎛ ⎞
⎢ ⎜ ⎟⎜ ⎟
⎢⎣ ⎝ ⎠⎝ ⎠
1 1 =12 thus 1=1⎤⎥
⎥⎦
Thus the first part of the induction (basis step)
is finished.
(b) Assuming the formula holds true for n , we will prove
it for n+1.
For the inductive step, we will assume
f f + f f + f f + ...+ f
f = f2
1 2 2 3 3 4
2n−1 2n 2n
And we will prove
f f + f f + f f + ...+ f
f +f f
+f
f
= f2
1 2 2 3 3 4
2n−1 2n 2n 2n+1 2n+1 2n+2 2n+2
Indeed,
f f + f f + f f + ...+ f
f +f f
+f
f
1 2 2 3 3 4
2n−1 2n 2n 2n+1 2n+1 2n+2
= f2 +f f
+f
f
2n 2n 2n+1 2n+1 2n+2
=f
⎛
⎜
⎜
⎝
f
2n 2n
+f
2n+1
⎞
⎟
⎟
⎠
+f
f
2n+1 2n+2
By inductive Hypothesis
=f
=f
=f
⎛
⎜
⎜
⎝
f
2n 2n
f
+f
2n 2n+2
⎛
⎜
⎜
⎝
f
+f
2n+2 2n
= f2
2n+2
2n+1
⎞
⎟
⎟
⎠
+f
f
2n+1 2n+2
f
2n+1 2n+2
+f
2n+1
⎞
⎟
⎟
⎠
By
f
2n
+f
2n+1
=f
2n+2
Thus the induction is complete, and I have proved
that the formula holds for all n.
QED
3. Linear Algebra (Matrices)
3.1 Cassini’s Identity:
n
⎛
⎞
2
− f n = ⎜⎝ − 1 ⎟⎠
f
f
n +1 n −1
⎡
⎢
⎢
⎢
⎢⎣
1 1 ⎤⎥
First we will introduce the transition matrix T =
⎥
1 0 ⎥⎥⎦
It is easy to see that this matrix satisfies
⎡
⎤
⎡
f n ⎤⎥
⎢ f
⎥
⎢
⎢ n+1⎥
⎢
⎥
for all n ≥1
⎢
⎥ =T ⎢
⎥
f ⎥
⎢ f
⎥
⎢
⎢⎣
n ⎥⎦ ⎣ n−1⎦
It is easy to prove (by induction) that
⎡
⎢
⎢
⎢
⎢
⎢
⎢⎣
f
T n = n+1
fn
for any n=1,2,3,…
fn
f
n−1
⎤
⎥
⎥
⎥
⎥
⎥
⎥⎦
Let’s take the determ inants of both sides:
det ⎡⎢⎣T n ⎤⎥⎦ = f
− f n2
n +1 n −1
f
But,
n
n
det ⎡⎢⎣T n ⎤⎥⎦ = ⎡⎢⎣ det T ⎤⎥⎦ = ⎝⎛⎜ − 1 ⎞⎟⎠
Therefore,
f
n
− f n2 = ⎛⎜⎝ − 1 ⎞⎟⎠
n +1 n −1
f
Thus we have proven the Cassini’s Identity.
Q ED
4. Combinatorics
4.1 0 - 1 Sequences
One important fact is that the number of 0 – 1 sequences
of length n without consecutive 1s is f
for every n ≥1.
n+2
Let’s prove this!
First denote by An the number of 0-1 sequences of
length n without consecutive 1s.
Here is an example of a string of length 8 without
consecutive 1s:
0 10 0 10 10
For n=1, we have a single
or a 1,
, so we have two possibilities a 0
Thus
A =2 , f
= f =2
1
1+2 3
holds true for n=1.
For n= 2 , we have two
01, 10
Thus A = 3 ,
2
for n= 2 .
f
therefore An = f
n+2
, so we have three possibilities 00,
= f =3
2+2 4
and An = f
holds true
n+2
Thus the first part of the induction (basis step) is finished.
I shall prove the induction step in the following way:
Assume A = f
k k +2
We want An = f
n+2
is true for all k < n.
n−1 cells
01001010
n cells
A
n−1
Ending with a “0”
01001010
n− 2 cells
01001001
Ending with a “1”
A
n−2
Therefore, An = A + A
=f
+f
= f + fn = f
n−1 n−2 n−1+2 n−2+2 n+1
n+2
QED
4.2 An Example of Combinatorial Proof:
We consider the identity f n2 + f 2 = f
n+1 2n+1
n−1
n−1
central cell
2⎛⎜⎝ n −1⎞⎟⎠ +1= 2n −1
Therefore, f
n−1
n−1
n− 2
n− 2
0
010
central cell is 0
central cell is 1
f
×f
= f2
n+1 n+1 n+1
= f 2 + f n2 .
2n+1 n+1
f n × f n = f n2
5. Educational Activity:
Many activities can be used in the classroom to
generate and investigate Fibonacci sequences. One is
to have students place 1 and 2 cent stamps across the
top of a postcard (facing with correct side up) in
different arrangements to make up certain postage
amounts. The number of different arrangements will
be a Fibonacci number.
END
:O)~