# 4.4 Pascalâ€™s Triangle

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```					4.4 Pascal’s Triangle
Pascal’s                Triangle

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
Each term is derived from the previous two terms above
it.
It is denoted  tn,r where tn,r ε Integers

n  row number
r  place number (counted from left side)

Counting always starts with Zero
Create the equation to form each entry
Tn,r = tn-1,r-1 + tn-1,r What does this mean?

Remember that we start counting using the 0!

Find the sum of the first 5 rows of the triangle
 2x

Predict the sum of row n
The first 6 terms in row 25 is 1,25,300,2300,12650 and 53130.
What are the first 6 terms in row 26?
Use the formula
T26,1 = t25,0 + t25,1
= 1 + 25
= 26

t26,2 = t25,1 + t25,2
= 25 + 300
= 325

t26,5 = t25,4 + t25,5
= 12650 + 53130
= 65780
Show the formula that would give you T34,8
Which row in Pascal’s triangle has the sum of its terms
equal to 32768?

From our investigations before we know that the sum of
each row is just 2n.
So we just need to find out for what value of n does
2n = 32768
 15
Divisibility
tn,2         Row 0  n/a      Therefore it is only
Is            ?
t n ,1       Row 1  n/a      divisible for odd-
Row 2  0.5 no   numbered rows.
Row 3  1 yes
Row 4  1.5 no
Row 5  2 yes
Row 6  2.5 no
Row 7  3 yes
Equilateral Triangles

1, 3, 6, 10, 15, 21, 28, 36, 45, 55

These are the number of “balls” that are placed in
an equilateral triangle

1
3            6
10
It follows the pattern 1+2+3+4+5+6+7+8
Look at Pascal’s Triangle….where do the numbers
fall that are in the pattern?

They fall on the diagonal with the term tn+1,2 The
term number is always 2, however the row is always
1 more than the number of columns in the equilateral.
Example
Find how many coins there are in 15 rows.
T15+1,2 =
t16,2
Perfect Squares
Is there a relationship between perfect squares and the sums of
pairs of entries in Pascal’s triangle?

n   n2   Entries in Pascal’s triangle Terms in the 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

Each perfect square greater than one is equal to the sum of a pair
of adjacent terms on the third diagonal of Pascal’s triangle
N2 = tn,2 + tn+1,2 for n>1
Homework
Pg 251 # 1, 2ace, 3bc, 4ac,5, 6, 8

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