FUN WITHSEQUENCES AND SERIES

TIM JEHL – MATH DUDE

THE PYRAMIDNumber 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 3DNumber 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

• Write an explicit formula for the perimeter of the nth square (Pn). 4L1(.707)n-1

• Find the formula for the nth partial sum of the perimeters (Sn) 4L1 (1-.707n)/(1-.707)

• If the series for the perimeters continues forever, what is the sum of the perimeters of all squares (S)? 4L1/(1-.707)

FIBONACCI

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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