1

Guide u5s2

1. To Built the Fibonacci series the rst thing to do is place the rst term

that is 1.

The next step is determinate the second element of the series, for this

we add the rst term of the series with the previous number of this, that is

to say, if the rst element is one then the previous number of this is zero,

therefore the second term of the series is 1 + 0 = 1.

In the same way to nd the third element of the series we add the second

element with the rst element of the series, that is to say, since the second

element of the series is 1 and the rst element of the series is 1 the third

element of the series is 1 + 1 = 2.

In the same way to nd the fourth element of the series we add the third

element of the series with second element of the series, since the third element

of the series is 2 and the second element is 1, therfore the fourth element is

2 + 1 = 3 and so on. T erm N ext term

F irst 1S econd term = 1 + 0 = 1

S econd 1T hird term =S econd

z}|{

1 + F irst

z}|{

1 = 2

T hird 2F ourth term =T hird

z}|{

2 + S econd

z}|{

1 = 3

F ourth 3F if th term =F ourth

z}|{

3 + T hird

z}|{

2 = 5

F if th 5S ixth term =F if th

z}|{

5 + F ourth

z}|{

3 = 8

.

.

. .

.

. .

.

.

Using this we can write the rst ten terms the Fibonacci sequence:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55,…

2. Sum of the rst ten terms of the Fibonacci sequence is

1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 = 143.

Division of the sum between 11, 143 / 11 = 13.

Since the division is exact the sum of the rst ten terms of the Fibonacci

sequence it is multiple of 11.

3. Let’s choose 3 and 4, following the procedure given in problem 1 we

2

have: T erm N ext term

F irst 3

S econd 4T hird term =S econd

z}|{

4 + F irst

z}|{

3 = 7

T hird 7F ourth term =T hird

z}|{

7 + S econd

z}|{

4 = 11

F ourth 11F if th term =F ourth

z}|{

11 + T hird

z}|{

7 = 18

F if th 18S ixth term =F if th

z}|{

18 + F ourth

z}|{

11 = 29

.

.

. .

.

. .

.

.

3, 4, 7, 11, 18, 29, 47, 76, 123, 199.

4. The sum of the sequence previous is given by:

3 + 4 + 7 + 11 + 18 + 29 + 47 + 76 + 123 + 199 = 517.

The division of the sum by 11 is: 517 / 11 = 47.

Since the division is exact the sum of the rst ten term of the Fibonacci-

like sequence is multiple of 11.

5. Hypothesis: The sum of the rst ten term of any Fibonacci-like is

multiple of 11.

6. Let’s choose 7 and 8, following the procedure given in problem 1 we

have: 7, 8, 15, 23, 38, 61, 99, 160, 259, 419.

The sum of the rst ten term of the sequence is:

7 + 8 + 15 + 23 + 38 + 99 + 160 + 259 + 419 = 1089.

The division of the sum by 11, 1089 / 11 = 99.

Since the sum of the rst ten term of the sequence is multiple of 11 the

hypothesis apply to this sequence (remember if the division of the sum of the

rst ten term of the sequence between 11 is exact, the sum of the rst ten

term is multiple of 11)