The array of numbers below is called Pascal's Triangle, it was discovered by a French mathematician, Blaise Pascal. In the triangle each term is equal to the sum of both numbers directly above it. The first and last number of each row is 1 considering the number before it always 1.

In the formula for Pascal's triangle rows r represented by n, and positions r. Both rows and positioning start with 0.

If tn,r represents the term in row n, position r, then tn,r= tn-1,r-1 + tn-1, r.

row0 1

row1 1 1

row2 1 2 1

row3 1 3 3 1

row4 1 4 6 4 1

row5 1 5 10 10 5 1

positions: t0,0

t1,0 t1,1

t2,0 t2,1 t2,2

t3,0 t3,1 t3,2 t3,3

t4,0 t4,1 t4,2 t4,3 t4,4

t5,0 t5,1 t5,2 t5,3 t5,4 t5,5

t6,0 t6,1 t6,2 t6,3 t6,4 t6,5 t6,6

Example:

tn,r=tn-1,r-1 +tn-1,r

=6,2

=,1+ t5,2

You can use the following chart to help you count position on rows

number of rows number of #'s in total term in Pascal's triangle

1 1 t2,2

2 3 t3,2

3 6 t4,2

4 10 t5,2

5 15 t6,2

6 21 t7,2

There are several number patterns in Pascal's Triangle

sum row (n)

1 1 1

1 + 1 * 2 2

1 + 2 + 1 4 3

1 + 3 + 3 + 1 8 4

1 + 4 + 6 + 4 + 1 16 5

1 + 5 + 10 + 10 + 5 + 1 32 6

1 + 6 + 15 + 20 + 15 + 6 + 1 64 7

The sum of the terms in row n of the triangle is n

2

Example A: 13

2 = 8192

Example B: which row has a sum of 128

using your calculator, 128= 2 to the power of 7.

Therefore the row which has the sum of 128 is row 7

log128 divided by log2 equals 7

Homework pg. 251 # 1-5, 7a, 9-11

Check My Work: (OSMAN OSMAN)

Click on the link below & open the file, it's waaaaay esier..!!

http://mdm4u1.wikidot.com/local--files/4-4-pascal-s-triangle/Chapter4.doc

Osman Osman

MDM4U0

Chapter 4 ( 4.4 Pascal's Triangle)

4.4 Pascal's Triangle:

Pascal's Triangle is an array of numbers arranged in staggered rows.

Check the diagram below:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

Pascal's method of building his triangle is simple , where each term equals the sum of the two terms immediately above it.

The first and the last row are both equal to 1 since the only term immediately above them t is always a 1.

This process or method of arranging these numbers is represented by a formula:

tn,r = t n-1, r-1 + tn-1, r

Where t n,r represents the term in a row n , position r .

For example: t6,2 = t5,1 + t5,2

Note that both the row and position labeling begin with 0 . (check diagram below)

1 Row 0 t 0,0

1 1 Row 1 t 1,0 t1,1

1 2 1 Row 2 t2,0 t2,1 t2,2

1 3 3 1 Row 3 t 3,0 t3,1 t3,2 t3,3

1 4 6 4 1 Row 4 t4,0 t4,1 t4,2 t4,3 t4,4

1 5 10 10 5 1 Row 5 t5,0 t5,1 t5,2 t5,3 t5,4 t5,5

1 6 15 20 15 6 1 Row 6 t6,0 t6,1 t6,2 t6,3 t6,4 t6,5 t6,6

Applications on Pascal’s Triangle:

Example 1: (Pascal’s Method)

a) The first 6 terms in row 25 of Pascal’s triangle are 1, 25 , 300 , 2300 , 12650 , and 53130. Determine the first 6 terms in row 26.

b) Use Pascal’s method to write a formula a for each of the following terms

i) t12,5

ii) t40,32

iii) tn+,r+1

Solution:

a) t26,1 = 1 t26,2 = 1+25 t26,3 = 25 +300 = 325

t26,4 = 300 + 2300 t26,5 = 2300 + 12650 t26,6 = 12650 + 53130

= 2600 = 14950 = 65780

b) i) t12,5 = t11,4 + t11,5

ii) t40,32 = t39,31 + t39,32

iii) tn+1,r+1 = tn,r + tn,r+1

Example 2: (Row Sums)

Which row in Pascal’s Triangle has the sum of its terms equal to 32768?

Solution:

The terms in any row n is 2^n . Dividing 32768 by 2 repeatedly, you find that 32768 = 2^15. Thus, it is row 15 of Pascal’s Triangle that has terms totalling 32768.

Example 3: (Divisibility)

Determine whether tn,2 is divisible by tn,1 in each row of Pascal’s Triangle

Solution:

Row tn,2 / tn,1 Divisible

0 and 1 n/a n/a

2 0.5 no

3 1 yes

4 1.5 no

5 2 yes

6 2.5 no

7 3 yes

It appears that tn,2 is divisible by tn,1 only in odd-numbered rows.

However, 2tn,2 is divisible by tn,1 in all rows that have three or more terms.

Example 4: (Triangular Numbers)

Coins can be arranged in the shape of an equilateral triangle as shown:

a) Continue the pattern to determine the number of coins in triangle with four, five , and six rows.

b) Locate these numbers in Pascal’s triangle.

c) Relate Pascal’s triangle to the number of coins in a triangle with n rows.

d) How many coins are in a triangle with 12 rows?

Solution:

a) The numbers of coins in the triangles follows the pattern 1+2+3+…… as shown in the table below.

b) The number of coins in the triangles matches the entries on the third diagonal of Pascal’s triangle.

- Row # of coins Term in Pascal’s Triangle

1 1 t2,2

2 3 t3,2

3 6 t4,2

4 10 t5,2

5 15 t5,2

6 21 t6,2

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1

c) If we compare the entries in the first and third columns of the table, we observe, the the number of the term from Pascal’s triangle is always one greater than the number of rows in the equilateral triangle. The position of the term in the row, r, is always 2. Thus, the number of coins in a triangle with n rows is equal to the term tn+1, 2 in Pascal’s triangle.

d) t12+1,2 = t13,2 = 78

A triangle with 12 rows contains 78 coins.

Hint:

Triangular Numbers : are numbers that correspond to the number of items stacked in a triangular array.

Example 5: (Perfect Squares)

Find a relationship between perfect squares and the sum of pairs of entries in Pascal’s triangle?

Solution:

Looking at the third diagonal in Pascal’s triangles.

n N^2 Entries in Pascal’s Triangle Term in Pascal’s Triangle

1 1 1 t2,2

2 4 1+3 t2,2 + t3,2

3 9 3+6 t3,2 + t4,2

4 16 6+10 t4,2 + t5,2

Key Concepts:

• Each term in Pascal’s Triangle is equal to the sumof two adjacent terms in the row immediately above tn,r = tn-1,r-1 + tn-1,r where tn,r represents the rth term in a row n.

• The sum of terms in Pascal’s triangle in row n in Pascal’s Triangle is 2^n.

• The terms in the third diagonal of Pascal’s triangle are triangular numbers.

Many other number patterns occur in Pascal’s triangle.

Usefull Sites for students who are interested in Pascal’s Traingle:

http://mathforum.org/workshops/usi/pascal/pascal/sierpinski.html

Applications of Pascal's Triangle to Sierpinski's triangle and fractals

http://mathforum.org/workshops/usi/pascal/pascal_hsdisc.html

Explorations in Pascal's triangle using number patterns

http://mathforum.org/workshops/usi/pascal/mid.color_pascal.html

Explorations in Pascal's Triangle involving Coloring Multiples

http://hsb.iitm.ernet.in/~jm/mar_april02/articles/pascal.htm

Background in Pascal's Triangle

http://www.roma.unisa.edu.au/07305/pascal.htm

Binomial Theorem and the Pascal Triangle

http://www.pbs.org/mathline

The Smithville Families - applications to probability

By: Osman Osman