Page 38 - PRE-U STPM MATHEMATICS (T) TERM 1

Mathematics Term 1 STPM Chapter 2 Sequences and Series
n
The expansion of (1 + x) , n  Z +
n
From the binomial theorem for (a + b) , when n is a positive integer with a = 1 and b = x, we have a finite series
n
n
n
n
n
2
n
r
(1 + x) = 1 + 1 2 x + 1 2 x + … + 1 2 x + … + 1 n –1 2 x n – 1 + x
2
1
r
n
r
= 1 + nx + n(n – 1) x + … + n(n – 1) … (n – r + 1) x + … + nx n – 1 + x .
2
2! r!
n
n
In general, if we want to expand (a + b) , it would be easier if we change it to the form (1 + x) as shown below:
3 1
n
(a + b) = a 1 + b 24 n
a
1
= a 1 + b 2 n
n
a
n
n
= a (1 + x) , where x = b . 2
a
Example 34
Use the binomial theorem to expand (1 + x) .
7
7
7
7
7
7
7
3
6
4
2
5
7
Solution: (1 + x) = 1 + 1 2 x + 1 2 x + 1 2 x + 1 2 x + 1 2 x + 1 2 x + x 7
5
1
2
4
6
3
4
3
6
= 1 + 7! x + 7! x + 7! x + 7! x + 7! x + 7! x + x 7
2
5
6! 1! 5! 2! 4! 3! 3! 4! 2! 5! 1! 6!
6
3
2
= 1 + 7x + 7 · 6 x + 7 · 6 · 5 x + 7 · 6 · 5 x + 7 · 6 x + 7x + x 7
4
5
1 · 2 1 · 2 · 3 1 · 2 · 3 1 · 2
3
= 1 + 7x + 21x + 35x + 35x + 21x + 7x + x 7
6
4
2
5
Example 35
4
Using the binomial theorem, find the expansion of (3x + 4y) .
4
4
4
Solution: (3x + 4y) = (3x) + 1 2 (3x) (4y) + 1 2 (3x) (4y) + 1 2 (3x)(4y) + (4y) 4
3
4
3
4
2
2
2
3
1
4 4
4 4
3
3
3
= 3 x + 4·3 ·4x y + 6·3 ·4 x y + 4·3·4 xy + 4 y
2 2 2 2
3
3
4
3
= 81x + 432x y + 864x y + 768xy + 256y 4
2 2
Example 36
4
15
Find the coefficient of x in the expansion of (2x – 1) .
Solution: From the binomial theorem,
n n
n
b
(a + b) = ∑ 1 2 a n – r r
r = 0 r
n
th
b
where the (r + 1) term = 1 2 a n – r r
r
15 15
15
Thus (2x – 1) = ∑ 1 2 (2x) 15 – r (–1) r
r = 0 r
4
The term in x is when 15 – r = 4, i.e. r = 11.
15
Coefficient of x = 1 2 2 (–1) 11
4
4
11
= – 15 × 14 × 13 × 12 × 2 4
1 × 2 × 3 × 4
= –21 840

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02 STPM Math T T1.indd 125 3/28/18 4:21 PM