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Fibonacci Sequence and Its Properties, Assignments of Mathematics

The University of Wyoming (UW)Mathematics

Mathematical proofs for the properties of the fibonacci sequence, including the gcd of 1 and the nth fibonacci number for all positive integers n.

Typology: Assignments

Pre 2010

Uploaded on 08/19/2009

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Math 3200 Homework #9
Due Monday February 23, 2004
Exercise Set 3
41. Find the fifth term in the expansion of
11
5
6
5
2
x
x
.
The fifth term in the expansion of
11
5
6
5
2
x
x
is:
22
2285
20
4
427
4
5
7
6
000,400,26
23511
5
2
1234
8910115
2
4
11
x
x
x
x
x
x
43. Proposition. If
p
is a prime, then
ppp
baba )(
(mod
)p
.
Proof. By Corollary 2.4.4
baba
p
)(
(mod
)p
, and
baba
pp
(mod
)p
. So
ppp
baba )(
(mod
)p
.
45. Proposition. Let
be a positive integer, then
n
n
nnn 2
10
.
Proof 1. Consider
n
11
. Clearly
n
n
211
. On the other hand by applying the
binomial theorem we have
.
10
111
2
11
1
1
0
11
221
n
nnn
n
nnnn
nnnn
n
Thus
n
n
nnn 2
10
.
Proof 2. We proceed by induction on
. The statement is true when
1n
since
1
211
1
1
0
1
. We assume that the statement is true for
kn
, that is
k
k
kkk 2
10
. Now consider
1
11
1
1
2
1
1
1
0
1
k
k
k
k
k
kkkk
pf3

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Download Fibonacci Sequence and Its Properties and more Assignments Mathematics in PDF only on Docsity!

Math 3200 Homework

Due Monday February 23, 2004

Exercise Set 3

41. Find the fifth term in the expansion of

11

5

6

x

x.

The fifth term in the expansion of

11

5

6

x

x is:

22

5 8 22

20

4

7 42

4

5

7 6

x

x

x

x

x

x

^ 

43. Proposition. If

p

is a prime, then

p p p

( a  b )  a  b (mod p^ ).

Proof. By Corollary 2.4.4 a^ b a b

p

(  )   (mod p ), and

a b a b

p p

   (mod^

p ). So

p p p

( a  b )  a  b (mod p ) . 

45. Proposition. Let

n

be a positive integer, then

n

n

n n n

2

0 1

 

  

 

Proof 1. Consider  

n

1  1. Clearly  

n (^) n

1  1  2. On the other hand by applying the

binomial theorem we have

1 2 2

 

n

n n n

n

n n n n n n n n n

Thus

n

n

n n n

2

0 1

 

  

 

Proof 2. We proceed by induction on n^. The statement is true when n  1 since

1 1 1 2

1

1

0

1

   

 

. We assume that the statement is true for n  k , that is

k

k

k k k

2

0 1

 

  

 

. Now consider

 

   

  

   

   

 

1

1 1

1

1

2

1

1

1

0

1

k

k

k

k

k

k k k k

 

 

 

 

  

 

 

 

 

 

1

1

0 0 1 1 2 2 1 1

1

k

k

k

k

k

k

k

k

k

k k k k k k

 

 

 

 

  

 

 

 

 

k

k

k

k

k

k

k

k

k

k k k k k k

0 0 1 1 2 2 1 1

k

k

k

k

k

k k k k

k  22

1

k

Thus when the statement is true for n  k it is also true for n  k  1. Hence by the

principle of mathematical induction the statement is true for all integers n^. 

59. Let

1 2 3

f f f

be the Fibonacci sequence, defined by 1 ,^1

1 2

f  f  , and

 1  2

n n n

f f f

for n^ ^3. The first 15 numbers in the Fibonacci sequence are:

60. Proposition. Let f^ 1 ,^ f 2 , f 3 ,be the Fibonacci sequence, defined by 1 ,^1

1 2

f  f  ,

and f^ n ^ fn  1  fn  2 for n  3. Then GCD ^ ,^ ^1

1

n n

f f for all positive integers n .

Proof 1. We proceed by induction on

n

. The statement is clearly true for n^ ^1. We

assume that the statement is true for n  k , that is GCD ^ ,^ ^1

1

k k

f f. Now consider

GCD ^ 

1 2

kk

f f. Let d be a nonnegative common divisor of

k  1

f

and k  2

f

. Since

k  2  3 ,

 2  1

k k k

f f f

. So 1

k

d f

and 1

k k

d f f

, and it follows from Proposition

1.1.2 (ii) that k

d | f

. Now we know that GCD

1

k k

f f

so d^ ^1. We have shown

that the only nonnegative common divisors of f^ k  1 and f^ k  2 is 1, hence GCD

1 2

kk

f f. Thus when the statement is true for n  k then it is also true for

n  k  1. So by the principle of mathematical induction the statement is true for all

positive integers n^. 

Proof 2. We proceed by induction on

n

. The statement is clearly true for n^ ^1. We

assume that the statement is true for n  k , that is GCD ^ ,^ ^1

1

k k

f f. Now consider

GCD ^ 

1 2

kk

f f. Since k  2  3 ,

1 2

 

k k k

f f f

. So by Proposition 1.2.2 it

follows that GCD

 1  2

k k

f f

GCD

1

k k

f f

. Thus when the statement is true for

n  k then it is also true for n  k  1. So by the principle of mathematical induction the

statement is true for all positive integers n^. 

Proof 3. We proceed by induction on

n

. The statement is clearly true for n^ ^1. We

assume that the statement is true for n  k , that is GCD ^ ,^ ^1

1

k k

f f. Then there exist