THE BINOMIAL THEOREM
PROBLEM SOLVING
In this section we consider binomial expansions of the form
( a + b ) n ( c + d ) m .
Example 1 Expand ( 𝑥 2 + 2 ) (𝑥 +
1 𝑥
)
5 and collect the like terms.
Solution
First, we use the Binomial Theorem to find the 6 terms in the expansion of
(𝑥 +
1
) 𝑥
5
. Next, each of the 6 tuerms is multiplied by x 2 and by 2. Finally, we add the like terms.
(
(𝑥 +
1 𝑥
)
5
= (
5
0
) 𝑥
(
5
4
5
) 𝑥 (
+ (
5
1
1 𝑥
)
4
= 𝑥 5 + 5𝑥 3
) 𝑥 4 (
1 𝑥
) + (
5
2
) 𝑥
+ (
5
5
) (
1 𝑥
)
5
+ 10𝑥 +
10 𝑥
+ 𝑥
5
3
3
+
(
1 𝑥 𝑥
1
5
)
2 𝑥 2 + 2 ) (𝑥 +
1 𝑥
)
5
= ( 𝑥 2 + 2 ) (𝑥 5 + 5𝑥 3
+ (
+ 10𝑥 +
5
3
10 𝑥
) 𝑥 2
5
+ 𝑥 3
(
1 𝑥
)
3
+
1
+ 𝑥 5
)
= 𝑥 7 + 5𝑥
+ 2𝑥
5
5
+ 10𝑥
+ 10𝑥
3
3
5
+ 10𝑥 +
+ 20𝑥 + 𝑥
20
+ 𝑥
1
+ 𝑥 3
10 𝑥 3
+
2 𝑥 5
= 𝑥 7 + 7𝑥 5 + 20𝑥 3 + 30𝑥 +
25 𝑥
+
11 𝑥 3
+
2 𝑥 5

Example 2 Find the first three terms in the expansion (𝑥 + 1) 5 (𝑥 − 2) 6
.
Solution
We use the Binomial Theorem to expand each of the binomials in the question. Since we are interested only in the first 3 terms, we need to find, at most, 3 terms in the expansion of these binomials.
(𝑥 + 1) 5 = (
5
= 𝑥
0
5
) 𝑥 5
+ 5𝑥
+ (
5
4
) 𝑥 4
1
+ 10𝑥
(1) + (
3 + …
5
2
) 𝑥 3 (1) 2 + …
(𝑥 − 2) 5 = (
6
= 𝑥
0
6
) 𝑥 6 + (
− 12𝑥 5
6
1
) 𝑥 5 (−2) + (
+ 60𝑥 4 − …
6
2
) 𝑥 4 (−2) 2
(𝑥 + 1) 5 (𝑥 − 2) 6 = (𝑥 5 + 5𝑥 4 + 10𝑥 3 + … )(𝑥
= 𝑥 11 − 12𝑥 10
+ 5𝑥 10
+ 60𝑥 9
− 60𝑥 9
− …
+ 300𝑥
6
8
= 𝑥 11 − 7𝑥 10
+ 10𝑥
+ 10𝑥
9
9
− 120𝑥 8
+ …
+ …
− 12𝑥
− …
5
+ 600𝑥
+ 60𝑥
7 − …
4 − … )

Example 3 In the expansion of ( 𝑥 2 – 1) 8 ( x + 2) 7 , find the term containing x 8 .
Solution
First, we find the general term in the expansion of each binomial.
For (𝑥 2 − 1) 8
, the general term is 𝑡 𝑟+1
= (
8 𝑟
) (𝑥 2 ) 8−𝑟 (−1) 𝑟
= (
8 𝑟
) (−1) 𝑟 𝑥 16 −2𝑟 , 0 ≤ 𝑟 ≤ 8
For (𝑥 + 2) 7
, the general term is 𝑡 𝑠 +1
= (
7 𝑠
) 𝑥 7 −𝑠 (2 𝑠 ) , 0 ≤ 𝑠 ≤ 7
Note: We use different variables, r and s , for each general term.
The unsimplified term in the expansion of (𝑥 2 calculated by multiplying each term of (𝑥 2 − 1) 8
− 1) 8 (𝑥 + 2) with each term of
7
are
(𝑥 +
2) 7
.
Therefore, each term is of the form
[(
8 𝑟
) (−1) 𝑟 𝑥 16−2𝑟 ] [(
7 𝑠
) (2 𝑠 )𝑥 7−𝑠 ] = (
8 𝑟
) (
7 𝑠
) (−1) 𝑟 (2 𝑠 ) 𝑥 23−2𝑟−𝑠
.
For the terms containing 𝑥 8
,
23 – 2 r – s = 8 → 2 r + s = 15
We look for integral solutions to this equation, where 0 ≤ 𝑟 ≤ 8 and 0 ≤ 𝑠 ≤ 7 . By trial and error we find four solutions:
r = 4 and s = 7
r = 5 and s = 5
r = 6 and s = 3
r = 7 and s = 1

Therefore, in the unsimplified expansion (𝑥 2 four terms containing 𝑥 8
.
− 1) 8 (𝑥 + 2) 7 of there are
When r = 4 and s = 7,
(
8 𝑟
) (
7 𝑠
) (−1) 𝑟 (2 𝑠 ) 𝑥 23−2𝑟−𝑠 = (
8
4
) (
= 8960
7
7 𝑥
) (−1)
8
4 (2
= 70(1)(1)(128) 𝑥
7
8
) 𝑥 8
When r = 5 and s = 5,
(
8 𝑟
) (
7 𝑠
) (−1) 𝑟 (2 𝑠 ) 𝑥 23−2𝑟−𝑠 = (
8
5
) (
7
5
) (−1) 5 (2 5 )
= 56(21)(−1)(32) 𝑥 8 𝑥 8
= −37632 𝑥 8
When r = 6 and s = 3,
(
8 𝑟
) (
7 𝑠
) (−1) 𝑟 (2 𝑠 ) 𝑥 23−2𝑟−𝑠 = (
8
6
) (
7
3
) (−1) 6 (2
= 28(35)(1)(8) 𝑥 8
3 ) 𝑥 8
= 7840 𝑥 8
When r = 7 and s = 1,
(
8 𝑟
) (
7 𝑠
) (−1) 𝑟 (2 𝑠 ) 𝑥 23−2𝑟−𝑠 = (
8
7
) (
7
1
) (−1) 7
= 8(7)(−1)(2) 𝑥
= −112 𝑥 8
(2 1
8
) 𝑥 8
The term containing 𝑥 8
is the sum of these four terms, – 20944 𝑥 8
.

Example 3
Use the Binomial Theorem and the identity (1 + 𝑥) 𝑛 (1 + 𝑥) 𝑛 to prove
= (1 + 𝑥) 2𝑛
( 𝑛
0
)
2
+ ( 𝑛
1
)
2
+ ( 𝑛
2
)
2
+ ⋯ + ( 𝑛 𝑟
) 𝑟
+ ⋯ + ( 𝑛 𝑛
)
2
= (
2𝑛 𝑛
) .
In sigma notation, 𝑛
∑ ( 𝑛 𝑟 𝑟=0
)
2
= (
2𝑛 𝑛
) .
Solution
We find the coefficient of
(1 + 𝑥) 𝑛 (1 + 𝑥) 𝑛 𝑥 𝑛 in the expansion of each side of the identity
= (1 + 𝑥) 2𝑛 .
In the expansion of (1 + 𝑥) 2𝑛 , the general term is 𝑡 𝑟+1
= (
2𝑛 𝑟
) (1 2𝑛−𝑟 )𝑥 𝑟
= (
2𝑛 𝑟
) 𝑥 𝑟
For the term containing 𝑥 𝑛
, r = n , and the coefficient of this term is (
2𝑛 𝑛
) .

The general term in the expansion of the first (1 + 𝑥) 𝑛
is 𝑡 𝑟+1
= ( 𝑛 𝑟
) (1 𝑛−𝑟 )(𝑥 𝑟 )
= ( 𝑛 𝑟
) 𝑥 𝑟
The general term in the expansion of the second (1 + 𝑥) 𝑛
is 𝑡 𝑠+1
= ( 𝑛 𝑠
) (1 𝑛−𝑠 )(𝑥 𝑠 )
= ( 𝑛 𝑠
) 𝑥 𝑠
Each term in the expansion of (1 + 𝑥) 𝑛 (1 + 𝑥) 𝑛
will be of the form
( 𝑛 𝑟
) 𝑥 𝑟 ( 𝑛 𝑠
) 𝑥 𝑠
or ( 𝑛 𝑟
) ( 𝑛 𝑠
) 𝑥 𝑟 + 𝑠
.
For the terms containing 𝑥 𝑛
, r + s = n , 0 ≤ r , s ≤ n .
There are n + 1 solutions to this equation. r = 0 and s = n r = 1 and s = n – 1 r = 2 and s = n – 2
… r = n and s = 0

Therefore, there are n + 1 terms containing 𝑥 𝑛
. Their coefficients are r = 0, s = n ( 𝑛 r = 1, s = n – 1 (
0 𝑛 r = 2, s = n – 2 (
1 𝑛
2
) ( 𝑛 𝑛
) (
) = ( 𝑛 𝑛 − 1 𝑛
) ( 𝑛 − 2 𝑛
0
)
2
) = ( 𝑛
1 𝑛
)
2
2
)
2
... ...
) = ( r = r , s = n – r ( 𝑛 𝑟
) ( 𝑛 𝑛 − 𝑟
… …
) = ( 𝑛 𝑟
)
2 r = n , s = 0 ( 𝑛 𝑛
) ( 𝑛
0
) = ( 𝑛 𝑛
)
2
Adding these, we obtain the coefficient of x n .
( 𝑛
0
)
2
+ ( 𝑛
1
)
2
+ ( 𝑛
2
)
2
+ ⋯ + ( 𝑛 𝑟
) 𝑟
+ ⋯ + ( 𝑛 𝑛
)
2
Since the expansion of (1 + 𝑥) 2𝑛
and (1 + 𝑥) 𝑛 (1 + 𝑥) 𝑛
are equal, the coefficients of 𝑥 𝑛
are the same. Therefore,
( 𝑛
0
)
2
+ ( 𝑛
1
)
2
+ ( 𝑛
2
)
2
+ ⋯ + ( 𝑛 𝑟
) 𝑟
+ ⋯ + ( 𝑛 𝑛
)
2
= (
2𝑛 𝑛
) .

EXERCISE
1. Expand ( x – 2) (2𝑥 + 1) 3
.
2.
Expand ( 𝑎
2
+ 1) (2𝑎 + 1) 4
.

3.
Expand ( x 2 + 1) (𝑥 −
1 𝑥
)
4
.
4. Find the first three terms in the expansion of ( x – 1) 7 ( x + 2) 9 .

5.
Find the first three terms in the expansion of (1 – x 2 ) 5 (1 + x 2 ) 10 .
6.
Find the first three terms in the expansion of ( x 2 – 1) 7 ( x 2 + 1) 8 .

7.
In the expansion of ( x 2 – 1) 6 ( x + 2) 9 , find the term containing x 5 .

8.
In the expansion of ( x 2 – 1) 5 (1 + x ) 8 , find the term containing x .

9.
In the expansion of ( a + 1) 7 ( a – 1) 7 , find the term containing a 8 .

10.
Expand ( a + b + c ) 4 . Hint: write a + b + c in the form a + ( b + c )

11.
Find the coefficient of x 5 in the expansion (1 + 2 x – x 2 ) 4 .

12. In a History class, there are m girls and n boys. A committee of k students is to be chosen, where k ≤ m and k ≤ n .
(a) In how many ways can this committee be chosen?
(b) How many of these committees contain no girls, one girl, two girls,
... , k girls?
(c) Write an equation relating parts (a) and (b).

(d) Use the Binomial Theorem and the identity (1 + 𝑥) 𝑚 (1 + 𝑥) 𝑛
(1 + 𝑥) 𝑚+𝑛
to prove
=
( 𝑚
0
) ( 𝑛 𝑘
) + ( 𝑚
1
) ( 𝑛 𝑘 − 1
) + ( 𝑚
2
) ( 𝑛 𝑘 − 2
) + ⋯ + ( 𝑚 𝑘
) ( 𝑛
0
) = ( 𝑚 + 𝑛 𝑘
) .
Hint: Determine the coefficient of 𝑥 𝑘 in the expansion of each side of the given identity.

ANSWERS
1. 8 x 4 – 4 x 3 – 18 x 2 – 11 x – 2
2. 8 a 5 + 32 a 4 + 44 a 3 + 28 a 2 + 8.5
a + 1
3. x 6 – 3 x 4 + 2 x 2 + 2 – 𝑥
3
2
+
1 𝑥 4
4.
x 16 + 11 x 15 + 39 x 14 + ...
5. 1 + 5 a 2 + 5 a 4 + ...
6.
x 30 + x 28 – 7 x 26 + ...
7. 4320 𝑥 5
8. – 8 x
9. – 35 𝑎 8
10. a 4 + 4 a 3 b + 4 a 3 c + 6 a 2 b 2 + 12 a 2 bc + 6 a 2 c 2 + 4 ab 3 + 12 ab 2 c
+ 12 abc 2 + 4 ac 3 + b 4 + 4 b 3 c + 6 b 2 c 2 + 4 bc 3 + c 4
11. – 8
12. (a) ( 𝑚 + 𝑛 𝑘
)
(b) ( 𝑚
0
) ( 𝑛 𝑘
) , ( 𝑚
1
) ( 𝑛 𝑘 − 1
) , ( 𝑚
2
) ( 𝑛 𝑘 − 2
) , ... , ( 𝑚 𝑘
) ( 𝑛
0
)
(c) ( 𝑚
0
) ( 𝑛 𝑘
) + ( 𝑚
1
) ( 𝑛 𝑘 − 1
) + ( 𝑚
2
) ( 𝑛 𝑘 − 2
) + ... + ( 𝑚 𝑘
) ( 𝑛
0
) = ( 𝑚 + 𝑛 𝑘
) or
∑ 𝑘 𝑟=0
( 𝑚 𝑟
) ( 𝑛 𝑘 − 𝑟
) = ( 𝑚 + 𝑛 𝑘
) .