Module 6, section B, question 1

The Fibonacci sequence is a sequence, in which, to get the next term you have to add the current term to the previous term.

eg. 1, 1, 2, 3, 5

In this article, we will be going over a series of challenges (The ones in Module 6, section B, Question 1)

Challenge 1:

The next 6 terms in the Fibonacci sequence are:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89

How to find the next terms:
I found the next term of the sequence by adding the current term with the previous term.
eg. to get the next term after 5, you do 3 + 5 = 8 2:

Challenge 2:

What happens when dividing each Fibonacci number by the one before it?
eg.
1 ÷ 1 = 1
2 ÷ 1 = 2
3 ÷ 2 = 1.5
…

When dividing each Fibonacci number by the one before it, the answer gets closer and closer to the golden ratio (Φ ≈ 1.618…).

1 ÷ 0 = undefined
1 ÷ 1 = 1
2 ÷ 1 = 2
3 ÷ 2 = 1.5
5 ÷ 3 = 1.˙6 (recurring)
8 ÷ 5 = 1.6
13 ÷ 8 = 1.625
21 ÷ 13 ≈ 1.615
34 ÷ 21 ≈ 1.619
55 ÷ 34 ≈ 1.6176
89 ÷ 55 = 1.6˙1˙8 (1 and 8 are recurring)
…

Notice that the bigger the numbers get, the closer the answer becomes to the golden ratio (Φ ≈ 1.618…).

Challenge 3:

What happens when dividing each Fibonacci number with the one after it?
eg.
0 ÷ 1 = undefined
1 ÷ 2 = 0.5
…

When dividing each Fibonacci number by the one after it, the answer gets closer to the reciprocal of the golden ratio which is, 1/Φ ≈ 0.618… . This value can be represented as lowercase Φ (φ).

0 ÷ 1 = undefined
1 ÷ 1 = 1
1 ÷ 2 = 0.5
2 ÷ 3 = 0.˙6 (recurring)
3 ÷ 5 = 0.6
5 ÷ 8 = 0.625
8 ÷ 13 ≈ 0.615
13 ÷ 21 ≈ 0.619
21 ÷ 34 ≈ 0.6176
35 ÷ 55 = 0.6˙1˙8 (1 and 8 are recurring)
…
Notice that as the number becomes bigger, the answer gets closer to φ ≈ 0.618… . This is the same as the reciprocal of the golden ratio (1/Φ)

Challenge 4:

Take three sets of three consecutive Fibonacci numbers - multiply the first and third and square the middle one. State what happens.
eg.
1, 1, 2
1 × 2 = 2
1² = 1
The difference is 1

1, 1, 2
1 × 2 = 2
1² = 1
2 - 1 = 1

1, 2, 3
1 × 3 = 3
2² = 4
3 - 4 = -1

2, 3, 5
2 × 5 = 10
3² = 9
10 - 9 = 1

3, 5, 8
3 × 8 = 24
5² = 24
24 - 25 = -1

5, 8, 13
5 × 13 = 65
8² = 64
65 - 64 = 1

8, 13, 21
8 × 21 = 168
13² = 169
168 - 169 = -1

13, 21, 34
13 × 34 = 442
21² = 421
442 - 441 = 1

21, 34, 55
21 × 55 = 1155
34² = 1156
1155 - 1156 = -1

Notice that the differences are always 1, -1

This is known as Cassini’s Identity which goes like this:

F stands for the Fibonacci sequence and, n stands for the number of the term.

lets take 3 consecutive Fibonacci numbers (1, 2, 3). The middle one is F₃ (the third term in the Fibonacci sequence), The one after it is F₃₊₁ = F₄ (The fourth term in the Fibonacci sequence). The one before is
F₃₋₁ = F₂ (The second term of the Fibonacci sequence).
We first multiply the other two numbers F₃₊₁ × F₃₋₁ (1 × 3 = 3) , and then square the middle one F₃² (2² = 4)
We then do F₃₊₁ × F₃₋₁ - F₃² = (-1)³. In this case n is 3 so we do (-1)³, which is -1.

Challenge 5:

add together the consecutive squares of Fibonacci numbers

1²
1² + 1²
1² + 1² + 2²
1² + 1² + 2² + 3²
1² + 1² + 2² + 3² + 5²
…

1² = 1
1² + 1² = 2
1² + 1² + 2² = 6
1² + 1² + 2² + 3² = 15
1² + 1² + 2² + 3² + 5² = 40
1² + 1² + 2² + 3² + 5² + 8² = 104
1² + 1² + 2² + 3² + 5² + 8² + 13² = 273
1² + 1² + 2² + 3² + 5² + 8² + 13² + 21² = 714
…
We then find the differences between the numbers
2 - 1 = 1
6 - 2 = 4
15 - 6 = 9
40 - 15 = 25
104 - 40 = 64
273 - 104 = 169
714 - 273 = 441

We then apply square roots to the answers
√1 = 1
√4 = 2
√9 = 3
√25 = 5
√64 = 8
√169 = 13
√441 = 21
The answers of the square roots will be a Fibonacci number
1, 2, 3, 5, 8, 13, 21…

Challenge 6:

This question is similar to the previous one
Find the products of consecutive Fibonacci numbers

eg.
1 × 1 = 1
2 × 1 = 2
2 × 3 = 6
3 × 5 =15
…

0 × 1 = 0
1 × 1 = 1
2 × 1 = 2
2 × 3 = **6
**3 × 5 =15
5 × 8 = 40
8 × 13 = 104
13 × 21 = 273
…

We then find the differences of the numbers
1 - 0 = 1
2 - 1 = 1
6 - 2 = 4
15 - 6 = 9
40 - 15 = 25
104 - 40 = 64
273 - 104 = 169

We then apply square root to the answers
√1 = 1
√1 = 1
√4 = 2
√9 = 3
√25 = 5
√64 = 8
√169 = 13

The answers of the square roots will always be Fibonacci numbers

Challenge 7:

The final challenge, Research where Fibonacci numbers are found in nature.

The Fibonacci sequence is found in many places in nature, like in flowers.

Some flowers, for example, will have a number of petals which correspond with the Fibonacci sequence.

Iris, Dwarf Lake (Iris lacustris) have 3 petals

buttercups have 5 petals

Most mountain Avens have 8 petals

Corn marigolds have 13 petals

some chicories have 21 petals