The binomial series 59
Hence
(a + x)
7
= a
7
+ 7a
6 x + 21a
5 x
2
+ 35a
4 x
3
+ 35a
3 x
4
+ 21a
2 x
5
+ 7ax
6
+ x
7
Problem 2. Determine, using Pascal’s triangle
method, the expansion of (2 p − 3q) 5 .
Comparing (2 p − 3q) 5 with (a + x) 5 shows that
a = 2 p and x = −3q.
Using Pascal’s triangle method:
(a + x)
5
= a
5
+ 5a
4 x + 10a
3 x
2
+ 10a
2 x
3
+ · · ·
Hence
(2 p − 3q)
5
= (2 p)
5
+ 5(2 p)
4
(−3q)
+ 10(2 p)
3
(−3q)
2
+ 10(2 p)
2
(−3q)
3
+ 5(2 p)(−3q)
4
+ (−3q)
5
i.e. (2p − 3q)
5
= 32p
5
− 240p
4 q + 720p
3 q
2
− 1080p
2 q
3
+ 810pq
4
− 243q
5
Now try the following exercise
Exercise 28 Further problems on Pascal’s
triangle
1. Use Pascal’s triangle to expand (x − y) 7 .
x 7 − 7x 6 y + 21x 5 y 2 − 35x 4 y 3
+ 35x 3 y 4 − 21x 2 y 5 + 7x y 6 − y 7
2. Expand (2a + 3b) 5 using Pascal’s triangle.
32a 5 + 240a 4 b + 720a 3 b 2
+ 1080a 2 b 3 + 810ab 4 + 243b 5
7.2 The binomial series
The binomial series or binomial theorem is a formula
for raising a binomial expression to any power without
lengthy multiplication. The general binomial expansion
of (a + x)
n is given by:
(a + x)
n
= a n + na n−1 x +
n(n − 1)
2!
a n−2 x 2
+
n(n − 1)(n − 2)
3!
a
n−3 x
3
+ · · ·
where 3! denotes 3 × 2 ×1 and is termed ‘factorial 3’.
With the binomial theorem n may be a fraction, a
decimal fraction or a positive or negative integer.
When n is a positive integer, the series is finite, i.e.,
it comes to an end; when n is a negative integer, or a
fraction, the series is infinite.
In the general expansion of (a + x) n it is noted that the
4th term is:
n(n − 1)(n − 2)
3!
a n−3 x 3 . The number 3 is
very evident in this expression.
For any term in a binomial expansion, say the r’th
term, (r − 1) is very evident. It may therefore be reasoned that the r’th term of the expansion (a + x)
n is:
n(n − 1)(n − 2). . . to (r − 1) terms
(r − 1)!
a
n−(r −1) x
r−1
If a = 1 in the binomial expansion of (a + x) n then:
(1 + x) n = 1 + nx +
n(n − 1)
2!
x 2
+
n(n − 1)(n− 2)
3!
x 3 +· · ·
which is valid for −1 < x < 1.
When x is small compared with 1 then:
(1 + x)
n
≈ 1 + nx
7.3 Worked problems on the
binomial series
Problem 3. Use the binomial series to determine
the expansion of (2 + x) 7 .
The binomial expansion is given by:
(a + x)
n
= a
n
+ na
n−1 x +
n(n − 1)
2!
a
n−2 x
2
+
n(n − 1)(n − 2)
3!
a
n−3 x
3
+ · · ·
Hence
(a + x)
7
= a
7
+ 7a
6 x + 21a
5 x
2
+ 35a
4 x
3
+ 35a
3 x
4
+ 21a
2 x
5
+ 7ax
6
+ x
7
Problem 2. Determine, using Pascal’s triangle
method, the expansion of (2 p − 3q) 5 .
Comparing (2 p − 3q) 5 with (a + x) 5 shows that
a = 2 p and x = −3q.
Using Pascal’s triangle method:
(a + x)
5
= a
5
+ 5a
4 x + 10a
3 x
2
+ 10a
2 x
3
+ · · ·
Hence
(2 p − 3q)
5
= (2 p)
5
+ 5(2 p)
4
(−3q)
+ 10(2 p)
3
(−3q)
2
+ 10(2 p)
2
(−3q)
3
+ 5(2 p)(−3q)
4
+ (−3q)
5
i.e. (2p − 3q)
5
= 32p
5
− 240p
4 q + 720p
3 q
2
− 1080p
2 q
3
+ 810pq
4
− 243q
5
Now try the following exercise
Exercise 28 Further problems on Pascal’s
triangle
1. Use Pascal’s triangle to expand (x − y) 7 .
x 7 − 7x 6 y + 21x 5 y 2 − 35x 4 y 3
+ 35x 3 y 4 − 21x 2 y 5 + 7x y 6 − y 7
2. Expand (2a + 3b) 5 using Pascal’s triangle.
32a 5 + 240a 4 b + 720a 3 b 2
+ 1080a 2 b 3 + 810ab 4 + 243b 5
7.2 The binomial series
The binomial series or binomial theorem is a formula
for raising a binomial expression to any power without
lengthy multiplication. The general binomial expansion
of (a + x)
n is given by:
(a + x)
n
= a n + na n−1 x +
n(n − 1)
2!
a n−2 x 2
+
n(n − 1)(n − 2)
3!
a
n−3 x
3
+ · · ·
where 3! denotes 3 × 2 ×1 and is termed ‘factorial 3’.
With the binomial theorem n may be a fraction, a
decimal fraction or a positive or negative integer.
When n is a positive integer, the series is finite, i.e.,
it comes to an end; when n is a negative integer, or a
fraction, the series is infinite.
In the general expansion of (a + x) n it is noted that the
4th term is:
n(n − 1)(n − 2)
3!
a n−3 x 3 . The number 3 is
very evident in this expression.
For any term in a binomial expansion, say the r’th
term, (r − 1) is very evident. It may therefore be reasoned that the r’th term of the expansion (a + x)
n is:
n(n − 1)(n − 2). . . to (r − 1) terms
(r − 1)!
a
n−(r −1) x
r−1
If a = 1 in the binomial expansion of (a + x) n then:
(1 + x) n = 1 + nx +
n(n − 1)
2!
x 2
+
n(n − 1)(n− 2)
3!
x 3 +· · ·
which is valid for −1 < x < 1.
When x is small compared with 1 then:
(1 + x)
n
≈ 1 + nx
7.3 Worked problems on the
binomial series
Problem 3. Use the binomial series to determine
the expansion of (2 + x) 7 .
The binomial expansion is given by:
(a + x)
n
= a
n
+ na
n−1 x +
n(n − 1)
2!
a
n−2 x
2
+
n(n − 1)(n − 2)
3!
a
n−3 x
3
+ · · ·