FUN WITH
SEQUENCES AND SERIES
TIM JEHL – MATH DUDE
THE PYRAMID
Number
of levels
Blocks in
last level
Blocks in
pyramid
1
1
1
2
2
3
3
3
6
4
4
10
5
5
15
6
6
21
7
7
28
8
8
36
PYRAMID MATH
• Arithmetic sequence for the number of blocks in each level
• an = n
• The total number of blocks in the pyramid is a series
• Sn = (n)(n+1)/2
THE PYRAMID IN 3D
Number
of levels
Blocks in
last level
Blocks in
pyramid
1
1
1
2
4
5
3
9
14
4
16
30
5
25
55
6
36
91
7
49
140
8
64
204
3D PYRAMID MATH
• Power sequence for the number of blocks in each level
• an = n2
• The total number of blocks in the pyramid is a series
• Sn = (n)(n+1)(2n+1)/6
NESTED SQUARES
NESTED SQUARES MATH
•
Complete the table below. The first square you drew corresponds to n = 1, the second square is n
= 2, etc.
•
•
•
•
Side Length of nth square (Ln) Ln = L1(.707)n-1
Side Length of nth square divided by 2 (Ln/2) (L1/2)(.707)n-1
Perimeter of nth square (Pn) 4L1(.707)n-1
nth partial sum of perimeters (Sn) 4L1 (1-.707n)/(1-.707)
•
•
Write a recursive formula for the perimeter of the nth square (Pn). Pn = .707Pn-1
•
•
Find the formula for the nth partial sum of the perimeters (Sn) 4L1 (1-.707n)/(1-.707)
Write an explicit formula for the perimeter of the nth square (Pn). 4L1(.707)n-1
If the series for the perimeters continues forever, what is the sum of the perimeters of all squares (S)? 4L1/(1-.707)
FIBONACCI
0
FIBONACCI MATH
• Write the recursive formula for the Fibonacci Sequence; you will need to specify the first two terms (1 and 1).
• Complete the following table, where fn is the nth term of the Fibonacci Sequence
• fn
• fn+1
• fn+1/fn
•
•
•
•
What value does fn+1 / fn approach as n gets bigger? This value is the golden ratio.
Take the golden ratio and subtract 1. Find the reciprocal of the golden ratio. Notice anything?
Take the golden ratio and add 1. Square the golden ratio. Notice anything? Pretty cool, huh?
Golden Ratio
SIERPENSKI’S TRIANGLE
SIERPENSKI’S TRIANGLE
• Complete the following table. Assume that your original triangle had an area of 100 cm2 and that n =1 is the removal of the first triangle.
•
•
•
•
•
Number of triangles removed during iteration (tn)
Area of one of the removed triangles (An)
Area removed during iteration (tn x An)
Total area remaining in Seirpenski’s Triangle
Total number of triangles removed (i.e. upside down triangles)
• Find a recursive formula for the area remaining in Seirpenski’s Triangle.
•
What is the area of Seirpenski’s Triangle after infinite iterations?
• Find a recursive formula for the number of upside down triangles in Seirpenski’s Triangle after n iterations.
• Number of triangles removed each time is geometric
•
an = 3n-1
• Area of the removed triangles is ¼ of the remaining area
• Area remaining is ¾ of the area - an = (3/4)n-1
SIERPENSKI’S TRIANGLE REVISITED
VON KOCH SNOWFLAKE
VON KOCH SNOWFLAKE MATH
• Complete the following table. Assume your first triangle had a perimeter of 9 inches.
• Number of line segments (tn)
• Length of each segment (Ln)
• Perimeter of snowflake (Pn)
•
•
•
•
•
•
Write a recursive formula for the number of segments in the snowflake (tn).
Write a recursive formula for the length of the segments (Ln).
Write a recursive formula for the perimeter of the snowflake (Pn).
Write the explicit formulas for tn, Ln, and Pn.
What is the perimeter of the infinite von Koch Snowflake?
Can you show why the area of the von Koch Snowflake is sum 4n-3x3.5/32n-7
PLUS
GOLDEN RATIO