THE BINOMIAL THEOREM
1. Binomial Expansions
A binomial is an expression of the form (a + b)n . Consider the expansion of these binomials for
various values of n.
(a + b)0 = 1 1
(a + b)1 = a+b 1 1
(a + b)2 = a 2 + 2ab + b 2 1 2 1
(a + b)3 = a 3 + 3a 2b + 3ab 2 + b3 1 3 3 1
(a + b)4 = a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4 1 4 6 4 1
Notice the powers of a descend as the powers of b ascend, and the sum of the powers is always
n.
The triangle of numbers is known as Pascal’s triangle : the sum of each pair of adjacent numbers
gives the number underneath the pair. The numbers in Pascal’s triangle correspond to the
coefficients in the binomial expansions.
Example 1 : Use Pascal’s triangle to expand (2 + 3x)5 .
We need the next row in the table, and this has coefficients 1, 5, 10, 10, 5, 1.
1 25 = 1 32 = 32
5 24 (3 x)1 = 5 16 3x = 240x
10 23 (3x)2 = 10 8 9x 2 = 720x 2
10 22 (3x)3 = 10 4 27x3 = 1080x 3
5 21 (3x)4 = 5 2 81x 4 = 810x 4
1 (3x)5 = 1 243x 5 = 243x5
And so (2 + 3x)5 = 32 + 240 x + 720 x 2 + 1080 x3 + 810 x 4 + 243x5 .
4
1
Example 2 : Use Pascal’s triangle to expand 5x − .
x
We need the row with coefficients 1, 4, 6, 4, 1.
1 (5 x)4 = 1 625x 4 = 625x 4
1 1
4 (5 x)3 − = 4 125x 3 − = −500x 2
x x
2
1 6 25x 2
1
−
2
6 (5 x) = = 150
x x2
3
1 4 5x −
1
−
20
4 (5 x)1 − =
x3
=
x2
x
4
1 1
1 1
1 − =
x4
=
x4
x
4
1 20 1
And so 5 x − = 625 x 4 − 500 x 2 + 150 − 2 + 4 .
x x x
, Example 3 : Find the coefficient of x 3 in the expansion of (7 − 2 x)5 .
Here we do not need to find the full expansion, just the term in x 3 .
1
5
10
10 72 (−2 x)3 = 10 49 −8x3 = −3920x 3
5
1
The coefficient of x is −3920.
3
Example 4 : Find the coefficient of x 5 in the expansion of (3x + 1)7 .
We need to continue Pascal’s triangle down a few rows.
1
7
21 (3x)5 12 = 21 243x5 1 = 5103x5
The coefficient of x 5 is 5103.
Example 5 : Find the coefficient of x 2 in the expansion of (1 − 2 x)(3x − 5)3 .
We first expand (3x − 5)3 …
1 (3x)3 = 1 27x 3 = 27x 3
3 (3x)2 (−5)1 = 3 9 x 2 −5 = −135x 2
3 (3 x)1 (−5)2 = 3 3x 25 = 225x
1 (−5)3 = 1 −125 = −125
And then multiply by (1 − 2 x) , only bothering to find those terms involving x 2 .
27x 3 −135x 2 225x −125
1 −135x 2
−2x −450x
The coefficient of x is −585.
2
Note that if we were far-sighted enough, we would only bother finding the terms in x and x 2 in
the expansion of (3x − 5)3 .
C2 p72 Ex 5A
1. Binomial Expansions
A binomial is an expression of the form (a + b)n . Consider the expansion of these binomials for
various values of n.
(a + b)0 = 1 1
(a + b)1 = a+b 1 1
(a + b)2 = a 2 + 2ab + b 2 1 2 1
(a + b)3 = a 3 + 3a 2b + 3ab 2 + b3 1 3 3 1
(a + b)4 = a 4 + 4a 3b + 6a 2b 2 + 4ab3 + b 4 1 4 6 4 1
Notice the powers of a descend as the powers of b ascend, and the sum of the powers is always
n.
The triangle of numbers is known as Pascal’s triangle : the sum of each pair of adjacent numbers
gives the number underneath the pair. The numbers in Pascal’s triangle correspond to the
coefficients in the binomial expansions.
Example 1 : Use Pascal’s triangle to expand (2 + 3x)5 .
We need the next row in the table, and this has coefficients 1, 5, 10, 10, 5, 1.
1 25 = 1 32 = 32
5 24 (3 x)1 = 5 16 3x = 240x
10 23 (3x)2 = 10 8 9x 2 = 720x 2
10 22 (3x)3 = 10 4 27x3 = 1080x 3
5 21 (3x)4 = 5 2 81x 4 = 810x 4
1 (3x)5 = 1 243x 5 = 243x5
And so (2 + 3x)5 = 32 + 240 x + 720 x 2 + 1080 x3 + 810 x 4 + 243x5 .
4
1
Example 2 : Use Pascal’s triangle to expand 5x − .
x
We need the row with coefficients 1, 4, 6, 4, 1.
1 (5 x)4 = 1 625x 4 = 625x 4
1 1
4 (5 x)3 − = 4 125x 3 − = −500x 2
x x
2
1 6 25x 2
1
−
2
6 (5 x) = = 150
x x2
3
1 4 5x −
1
−
20
4 (5 x)1 − =
x3
=
x2
x
4
1 1
1 1
1 − =
x4
=
x4
x
4
1 20 1
And so 5 x − = 625 x 4 − 500 x 2 + 150 − 2 + 4 .
x x x
, Example 3 : Find the coefficient of x 3 in the expansion of (7 − 2 x)5 .
Here we do not need to find the full expansion, just the term in x 3 .
1
5
10
10 72 (−2 x)3 = 10 49 −8x3 = −3920x 3
5
1
The coefficient of x is −3920.
3
Example 4 : Find the coefficient of x 5 in the expansion of (3x + 1)7 .
We need to continue Pascal’s triangle down a few rows.
1
7
21 (3x)5 12 = 21 243x5 1 = 5103x5
The coefficient of x 5 is 5103.
Example 5 : Find the coefficient of x 2 in the expansion of (1 − 2 x)(3x − 5)3 .
We first expand (3x − 5)3 …
1 (3x)3 = 1 27x 3 = 27x 3
3 (3x)2 (−5)1 = 3 9 x 2 −5 = −135x 2
3 (3 x)1 (−5)2 = 3 3x 25 = 225x
1 (−5)3 = 1 −125 = −125
And then multiply by (1 − 2 x) , only bothering to find those terms involving x 2 .
27x 3 −135x 2 225x −125
1 −135x 2
−2x −450x
The coefficient of x is −585.
2
Note that if we were far-sighted enough, we would only bother finding the terms in x and x 2 in
the expansion of (3x − 5)3 .
C2 p72 Ex 5A
