Although known to Chinese mathematicians many centuries earlier, the following pattern is named after the seventeenth century French mathematician Blaise Pascal. Pascal’s triangle contains many fascinating patterns. Each row from row 1 onwards begins and ends with ‘1’. Each other number along a row is formed by adding the two terms to its left and right from the preceding row. Row 0 1 Row 1 1 1 Row 2 1 2 1 Row 3 1 3 3 1 Row 4 1 4 6 4 1
The numbers in each row are called binomial coefficients. Interactivity
Pascal’s triangle int-2554
Expand (2x − 5)3.
Note: It is appropriate to use CAS technology to perform expansions such
as this.
THINK WRITE
1 Use the rule for expanding a perfect cube. (2x− 5)3
Using (a− b)3 = a3− 3a2b+ 3ab2− b3,
let 2x= a and 5 = b.
(2x− 5)3
= (2x)3 − 3(2x)2(5) + 3(2x)(5)2 − (5)3
2 Simplify each term. = 8x3 − 3 × 4x2 × 5 + 3 × 2x × 25 − 125
= 8x3 − 60x2 + 150x − 125
3 State the answer. ∴ (2x − 5)3 = 8x3− 60x2 + 150x − 125
4 An alternative approach to using the rule would be to write the expression in the form (a + b)3. (2x− 5)3 = (2x + (−5))3 Using (a+ b)3 = a3+ 3a2b+ 3ab2+ b3, let 2x= a and −5 = b. (2x− 5)3 = (2x + (−5))3 = (2x)3 + 3(2x)2(−5) + 3(2x)(−5)2 + (−5)3 = 8x3 − 60x2 + 150x − 125 Worked exAmple
6
The numbers 1,2,1 in row 2 are the coefficients of the terms in the expansion of (a+ b)2.
(a + b)2 = 1a2 +2ab +1b2
The numbers 1,3,3,1 in row 3 are the coefficients of the terms in the expansion of (a+ b)3.
(a + b)3 = 1a3 +3a2b+ 3ab2 +1b3
Each row of Pascal’s triangle contains the coefficients in the expansion of a power of (a+ b). Such expansions are called binomial expansions because of the two terms a and b in the brackets.
Row n contains the coefficients in the binomial expansion (a+ b)n.
To expand (a + b)4 we would use the binomial coefficients, 1,4,6,4,1, from row 4 to obtain:
(a + b)4 = 1a4 +4a3b+ 6a2b2 + 4ab3+ 1b4 = a4 + 4a3b+ 6a2b2 + 4ab3+ b4
Notice that the powers of a decrease by 1 as the powers of b increase by 1, with the sum of the powers of a and b always totalling 4 for each term in the expansion of (a+ b)4.
For the expansion of (a− b)4 the signs would alternate:
(a − b)4 = a4 − 4a3b+ 6a2b2 − 4ab3+ b4
By extending Pascal’s triangle, higher powers of such binomial expressions can be expanded.
Form the rule for the expansion of (a− b)5 and hence expand (2x− 1)5.
Note: It is appropriate to use CAS technology to perform expansions such
as this.
THINK WRITE
1 Choose the row in Pascal’s triangle which contains the required binomial coefficients.
For (a − b)5, the power of the binomial is 5. Therefore the
binomial coefficients are in row 5. The binomial coefficients are: 1, 5, 10, 10, 5, 1
2 Write down the required binomial expansion.
Alternate the signs:
(a − b)5 = a5− 5a4b + 10a3b5 − 10a2b3+ 5ab4 − b5 3 State the values to
substitute in place of a and b.
To expand (2x − 1)5, replace a by 2x and b by 1.
4 Write down the expansion. (2x − 1)5 = (2x)5− 5(2x)4(1) + 10(2x)3(1)2 − 10(2x)2(1)3 + 5(2x)(1)4 − (1)5 5 Evaluate the coefficients and state the answer. = 32x5 − 5 × 16x4 + 10 × 8x3− 10 × 4x2 + 10x − 1 = 32x5 − 80x4+ 80x3 − 40x2 + 10x − 1 ∴ (2x − 1)5= 32x5 − 80x4+ 80x3 − 40x2 + 10x − 1 Worked exAmple
7
Pascal’s triangle and binomial expansions
1 WE6 Expand (3x − 2)3. 2 Expand a
3 + b
2 3 and give the coefficient of a2b2.
3 WE7 Form the rule for the expansion of (a− b)6 and hence expand (2x − 1)6.
4 Expand (3x + 2y)4.
5 Expand the following.
a (3x+ 1)3 b (1 − 2x)3 c (5x+ 2y)3 d x 2 − y 3 3
6 Select the correct statement(s).
a (x+ 2)3= x3 + 6x2+ 12x + 8 b (x + 2)3 = x3+ 23
c (x+ 2)3= (x + 2)(x2− 2x + 4) d (x + 2)3 = (x + 2)(x2 + 2x + 4)
E (x+ 2)3= x3 + 3x2+ 3x + 8
7 Find the coefficient of x2 in the following expressions.
a (x+ 1)3− 3x(x + 2)2 b 3x2(x+ 5)(x − 5) + 4(5x − 3)3
c (x− 1)(x + 2)(x − 3) − (x − 1)3 d (2x2 − 3)3+ 2(4 − x2)3
8 a Write down the numbers in row 7 of Pascal’s triangle.
b Identify which row of Pascal’s triangle contains the binomial coefficients: 1, 9, 36, 84, 126, 126, 84, 36, 9, 1.
c Row 0 contains 1 term, row 1 contains 2 terms. How is the number of terms related to the row number of Pascal’s triangle?
9 Copy and complete the following table by making use of Pascal’s triangle.
Binomial
power Expansion Number of terms in the expansion Sum of indices in each term
(x+ a)2 (x+ a)3 (x+ a)4 (x+ a)5
10 Expand the following using the binomial coefficients from Pascal’s triangle.
a (x+ 4)5 b (x − 4)5
c (xy+ 2)5 d (3x − 5y)4
e (3− x2)4 f (1 + x)6 − (1 − x)6
11 a Expand and simplify (x− 1) + y 4.
b Find the term independent of x in the expansion of x 2 + 2x
6 .
c If the coefficient of x2y2 in the expansion of (x+ ay)4 is 3 times the coefficient of x2y3 in the expansion of (ax2 − y)4, find the value of a.
d Find the coefficient of x in the expansion of (1+ 2x)(1 − x)5. ExERcIsE 2.3
PRacTIsE
Work without cas
coNsolIdaTE
apply the most appropriate mathematical processes and tools
12 a The sum of the binomial coefficients in row 2 is 1 + 2 + 1 = 4. Form the sum of the binomial coefficients in each of rows 3 to 5.
b Create a formula for the sum of the binomial coefficients of row n. c Expand (1+ x)4.
d In the expansion in part c, let x = 1 and comment on the result.
e Using a suitably chosen value for x evaluate 1.14 using the expansion in part c.
13 a Expand (x+ 1)5 − (x + 1)4 and hence show that (x+ 1)5 − (x + 1)4 = x(x + 1)4.
b Prove (x+ 1)n+1 − (x + 1)n = x(x + 1)n.
14 A section of Pascal’s triangle is shown. Determine the values of a, b and c.
45 a
b 165 330
220 c
15 Pascal’s triangle can be written as: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1
a Describe the pattern in the second column.
b What would be the sixth entry in the third column?
c Describe the pattern of the terms in the third column by forming a rule for the
nth entry.
d What would be the rule for finding the nth entry of the fourth column?
16 Expand (1+ x + x2)4 and hence, using a suitably chosen value for x, evaluate 0.914.
The Yang Hui (Pascal’s) triangle as depicted in 1303 in a work by the chinese mathematician chu shih-chieh
MasTER
The binomial theorem
Note: The binomial theorem is not part of the Study Design but is included here to enhance understanding.
Pascal’s triangle is useful for expanding small powers of binomial terms. However, to obtain the coefficients required for expansions of higher powers, the triangle needs to be extensively extended. The binomial theorem provides the way around this limitation by providing a rule for the expansion of (x + y)n. Before this theorem can
be presented, some notation needs to be introduced.
factorial notation
In this and later chapters, calculations such as 7 × 6 × 5 × 4 × 3 × 2 × 1 will be encountered. Such expressions can be written in shorthand as 7! and are read as ‘7 factorial’. There is a factorial key on most calculators, but it is advisable to remember some small factorials by heart.
definition
n! = n × (n − 1) × (n − 2) × … × 3 × 2 × 1 for any natural number n. It is also necessary to define 0! = 1.
7! is equal to 5040. It can also be expressed in terms of other factorials such as: 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 7 × 6! or
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1
= 7 × 6 × 5!
This is useful when working with fractions containing factorials. For example: 7! 6! = 7 × 6! 6! = 7 or 5! 7!= 5! 7× 6 × 5! = 142
By writing the larger factorial in terms of the smaller factorial, the fractions were simplified.
Factorial notation is just an abbreviation so factorials cannot be combined arithmetically. For example, 3! − 2! ≠ 1!. This is verified by evaluating 3! − 2!.
3!− 2! = 3 × 2 × 1 − 2 × 1 = 6 − 2 = 4 ≠ 1
2.4
Interactivity The binomial theorem int-2555 Evaluate 5!− 3! + 50! 49! THINK WRITE1 Expand the two smaller factorials. 5! − 3! + 50! 49! = 5 × 4 × 3 × 2 × 1 − 3 × 2 × 1 + 50!49! Worked exAmple