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Chapter 6 Series
In this chapter you will learn how to:
■ ■ ■ ■
use the expansion of ( a + b ) n, where n is a positive integer recognise arithmetic and geometric progressions use the formulae for the nth term and for the sum of the first n terms to solve problems involving arithmetic or geometric progressions use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression.
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PREREQUISITE KNOWLEDGE Where it comes from
What you should be able to do
Check your skills
IGCSE / O Level Mathematics
Expand brackets.
1 Expand. a (2 x + 3)2
IGCSE / O Level Mathematics
Simplify indices.
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b (1 − 3x )(1 + 2 x − 3x 2 ) 2 Simplify.
a (5x 2 )3
b ( −2 x 3 )5
IGCSE / O Level Mathematics
3 Find the nth term of these linear sequences. a 5, 7, 9, 11, 13, …
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Find the nth term of a linear sequence.
b 8, 5, 2, −1, −4, …
FAST FORWARD
Why study series?
At IGCSE / O Level you learnt how to expand expressions such as (1 + x )2 . In this chapter we will learn how to expand expressions of the form (1 + x ) n where n can be any positive integer. Expansions of this type are called binomial expansions.
This chapter also covers arithmetic and geometric progressions. These are number sequences that have particular special properties. We will learn how to find the sum of the numbers in these progressions. Some fractal patterns can generate these types of sequences.
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6.1 Binomial expansion of (a + b)n Binomial means ‘two terms’.
The word is used in algebra for expressions such as x + 3 and 5x − 2 y.
You should already know that ( a + b )2 = a 2 + 2 ab + b2.
The expansion of ( a + b )2 can be used to expand ( a + b )3: ( a + b )3 = ( a + b )( a 2 + 2 ab + b2 )
= a 3 + 2 a 2b + ab2 + a 2b + 2 ab2 + b3
= a 3 + 3a 2b + 3ab2 + b3 Similarly, it can be shown that ( a + b )4 = a 4 + 4a 3b + 6a 2b2 + 4ab3 + b 4. Writing the expansions of ( a + b ) n out in order: ( a + b )0 = 1
( a + b )1 = 1a + 1b
1
1a + 1b
( a + b )2 = 1a 2 + 2 ab1a+2 1+b22 ab + 1b2
( a + b )3 = 1a 3 +1a33a 2+b3+a32bab+2 3+ab 1b23 + 1b3 ( a + b )4 = 1a 4 + 4a 3b + 6a 2b2 + 4ab3 + 1b 4
Original material © Cambridge University Press 2017
In the Pure Mathematics 2 and 3 Coursebook, Chapter X, you will learn how to expand these expressions for any rational value of n.
FAST FORWARD Properties of binomial expansions are also used in probability theory, which you will learn about if you go on to study the Probability and Statistics 1 Coursebook, Chapter X.
WEB LINK Explore the Sequences and Counting and Binomial stations on the Underground Mathematics website.
Chapter 6: Series
If you look at the expansion of ( a + b )4, you should notice that the powers of a and b form a pattern.
●
●
the first term is a 4 and then the power of a decreases by 1 whilst the power of b increases by 1 in each successive term all of the terms have a total index of 4 ( a 4, a 3b, a 2b2 , ab3 and b 4 ).
There is a similar pattern in the other expansions.
1 = 0: = 1: 1 1 = 2: 1 +2 1 = 3: 1 3+3 1 = 4: 1 4 6 4 1
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Each row always starts and finishes with a 1.
Each number is the sum of the two numbers in the row above it.
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n n n n n
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The coefficients also form a pattern that is known as Pascal’s triangle.
The next row is: n = 5 : 1 5 10 10 5 1
This row can then be used to write down the expansion of ( a + b )5: ( a + b )5 = 1a 5 + 5a 4b + 10 a 3b2 + 10 a 2b3 + 5ab 4 + 1b5
DID YOU KNOW?
EXPLORE 6.1
1
1
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1
1
1
1
2
3
1
3
1
1 10 10 5 1 1 6 15 20 15 6 1 1
4
6
4
5
There are many number patterns to be found in Pascal’s triangle. For example, the numbers 1, 4, 10 and 20 have been highlighted.
1
4
10
20
These numbers are called tetrahedral numbers.
1 What do you notice if you find the total of each row in Pascal’s triangle? 2 Can you use Pascal’s triangle to make the Fibonacci sequence (1, 1, 2, 3, 5, 8, 13, …)? 3 Pascal’s triangle has many other number patterns. Which number patterns can you find?
Original material © Cambridge University Press 2017
Pascal’s triangle is named after the French mathematician Blaise Pascal (1623–1662). Blaise Pascal is most famous for developing the first computing machine before he was 20 years old.
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WORKED EXAMPLE 6.1 Use Pascal’s triangle to find the expansion of: a (3x + 2)3
b (5 − 2 x )4
Answer a (3x + 2)3
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The index is 3 so use the row for n = 3 in Pascal’s triangle. (1, 3, 3, 1) (3x + 2)3 = 1(3x )3 + 3(3x )2 (2) + 3(3x )(2)2 + 1(2)3 b (5 − 2x )
4
= 27 x 3 + 54x 2 + 36x + 8
The index is 4 so use the row for n = 4 in Pascal’s triangle. (1, 4, 6, 4, 1)
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(5 − 2 x )4 = 1(5)4 + 4(5)3 ( −2 x ) + 6(5)2 ( −2 x )2 + 4(5)( −2 x )3 + 1( −2 x )4 = 625 − 1000 x + 600 x 2 − 160 x 3 + 16x 4
WORKED EXAMPLE 6.2
a Use Pascal’s triangle to expand (1 − 2 x )5.
b Find the coefficient of x 3 in the expansion of (3 + 5x )(1 − 2 x )5. Answer
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a (1 − 2 x )5
The index is 5 so the row for n = 5 in Pascal’s triangle. (1, 5, 10, 10, 5, 1)
(1 − 2 x )5 = 1(1)5 + 5(1)4 ( −2 x ) + 10(1)3 ( −2 x )2 + 10(1)2 ( −2 x )3 + 5(1)( −2 x )4 + 1( −2 x )5 = 1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5
b (3 + 5x )(1 − 2 x )5 = (3 + 5x )(1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5 ) The term in x 3 comes from the products:
(3 + 5x )(1 − 10 x + 40 x 2 − 80 x 3 + 80 x 4 − 32 x5 )
3 × ( −80 x 3 ) = −240 x 3 and 5x × 40 x 2 = 200 x 3
Coefficient of x 3 = −240 + 200 = −40
EXERCISE 6A
1 Use Pascal’s triangle to find the expansions of: a ( x + 2)3
b (1 − x )4
e ( x − y )4
f
(2 x + 3 y )3
c ( x + y )3
d (2 − x )3
g (2 x − 3)4
3 h x 2 + 2x3
Original material © Cambridge University Press 2017
3
Chapter 6: Series
2 Find the coefficient of x 3 in the expansions of: a ( x + 3)4
b (1 + x )5
e ( x − 2)5
f
(2 x − 1)4
c (3 − x )5
d (4 + x )4
g (4x + 3)4
x h 5 − 2
4
3 (3 + x )5 + (3 − x )5 = A + Bx 2 + Cx 4
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Find the value of A, the value of B and the value of C . 4 The coefficient of x 2 in the expansion of (3 + ax )4 is 216. Find the possible values of the constant a. 5 a Expand (2 + x )4 .
6 a Expand (1 + x )3 .
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b Use your answer to part a to express (2 + 3 )4 in the form a + b 3 .
b Use your answer to part a to express: i
(1 + 5 )3 in the form a + b 5
ii (1 − 5 )3 in the form c + d 5 .
c Use your answers to part b to simplify (1 + 5 )3 + (1 − 5 )3. 7 Expand (1 + x )(2 + 3x )4. 8 a Expand ( x 2 − 1)4 .
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b Find the coefficient of x 6 in the expansion of (1 − 2 x 2 )( x 2 − 1)4. 4
2 9 Find the coefficient of x 2 in the expansion of 3x − . x 4
3 10 Find the term independent of x in the expansion of x 2 − 2 . x
11 a Find the first three terms, in ascending powers of y, in the expansion of (1 + y )4 . b By replacing y with 5x − 2 x 2, find the coefficient of x 2 in the expansion of (1 + 5x − 2 x 2 )4.
12 The coefficient of x 2 in the expansion of (1 + ax )4 is 30 times the coefficient of x in the expansion of 3
1 + ax . Find the value of a. 3
4
1 13 Find the power of x that has the greatest coefficient in the expansion of 3x 4 − . x
14 a Write down the expansion of ( x + y )5 .
5
1 b Without using a calculator and using your result to part a, find the value of 10 correct to 4
the nearest hundred.
4
4
1 1 q 15 a Given that x 2 + − x 2 − = px5 + , find the value of p and the value of q. x x c 4 4 1 1 −2− . b Hence, without using a calculator, find the exact value of 2 + 2 2
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1 x 1 a Express x 3 + 3 in terms of y. x 1 b Express x5 + 5 in terms of y. x
16 y = x +
6.2 Binomial coefficients
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Pascal’s triangle can be used to expand ( a + b ) n for any positive integer n, but if n is large it can take a long time to write out all the rows in the triangle. Hence, we need a more efficient method to find the coefficients in the expansions. The coefficients in a binomial expansion are also known as binomial coefficients.
EXPLORE 6.2 Consider the expansion:
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(1+ x )5 = 1+ 5x + 10x 2 + 10x 3 + 5x 4 + x 5 The coefficients are: 1 5
10
10
5
1
Find the n C r function on your calculator. On some calculators this may be n C r n or . r
1 Use your calculator to find the value of:
5 5 5 5 5 5 , , , , and . 5 0 1 2 3 4
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2 What do you notice about your answers to question 1? 3 Complete the following four statements: 5 The coefficient of x 2 in the expansion of (1 + x )5 is . … … The coefficient of x r in the expansion of (1 + x ) n is . …
5 5 The coefficient of the 4th term in the expansion of ( 1 + x ) is . … … n The coefficient of the ( r + 1)th term in the expansion of ( 1 + x ) is . …
We write the binomial expansion of (1 + x ) n , where n is a positive integer. This is known as the Binomial theorem.
n n n n (1 + x ) n = + x + x 2 + … + x n 0 1 2 n
Original material © Cambridge University Press 2017
5 To find key in 0 5 nCr 0
Chapter 6: Series
The formula for the ( r + 1)th term in the expansion of (1 + x ) n is: n ( r + 1)th term = x r r
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The result can be extended to: n n n n ( a + b ) n = a n + a n −1b1 + a n − 2b2 + … + b n 1 2 0 n
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The formula for the ( r + 1)th term in the expansion of ( a + b ) n is: n ( r + 1)th term = a n − r b r r
WORKED EXAMPLE 6.3
Find, in ascending powers of x, the first four terms in the expansion of: a (1 + x )15 Answer
b (2 − 3x )20
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a (1 + x )15 = 15 + 15 x + 15 x 2 + 15 x 3 + … 2 0 1 3 = 1 + 15x + 105x 2 + 455x 3 + …
b (2 − 3x )10 = 10 210 + 10 29 ( −3x )1 + 10 28 ( −3x )2 + 10 27 ( −3x )3 + … 0 3 1 2 = 1024 − 15 360 x + 103 680 x 2 − 414 720 x 3 + …
You should also know how to work out the binomial coefficients without using a calculator. 5 5 From Pascal’s triangle, we know that = 1 and = 1. 0 5 In general, we can write this as:
n n 0 = 1 and n = 1
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5 5 5 5 We write , , and as: 1 2 3 4 5 5 = =5 1 1
5 5×4 = 10 = 2 2 ×1
5 5×4×3 = 10 = 3 3× 2 ×1
5 5×4×3×2 =5 = 4 4× 3× 2 ×1
n n × ( n − 1) × ( n − 2) × … × ( n − r + 1) r = r × ( r − 1) × ( r − 2) × … × 3 × 2 × 1
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WORKED EXAMPLE 6.4
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In general, if r is a positive integer less than n then:
8 a Without using a calculator, find the value of . 4 n b Find an expression, in terms of n, for . 4 Answer
n n × ( n − 1) × ( n − 2) × ( n − 3) n( n − 1)( n − 2)( n − 3) = b = 4×3×2 ×1 24 4
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8 8×7×6×5 = 70 a = 4×3×2 ×1 4
WORKED EXAMPLE 6.5 x When 1 − 3
n
is expanded in ascending powers of x, the coefficient of x 2 is 4. Find the value of n.
Answer
2 n x n × ( n − 1) x 2 n × ( n − 1) 2 Term in x 2 = − = x × = 3 2 1 9 18 × 2
n × ( n − 1) =4 18 n( n − 1) = 72
n2 − n − 72 = 0
( n − 9)( n + 8) = 0 n = 9 or n = −8 n=9
n must be a positive integer.
Original material © Cambridge University Press 2017
Chapter 6: Series
Checklist of learning and understanding Binomial expansions n Binomial coefficients, denoted by n C r or , can be found by using: r ●● Pascal’s triangle n n n! n × ( n − 1) × ( n − 2) × … × ( n − r + 1) the formulae = or = r × ( r − 1) × ( r − 2) × … × 3 × 2 × 1 r r !( n − r ) ! r
If n is a positive integer, the Binomial theorem states that:
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n n n n n (1 + x ) n = + x + x 2 + … + x n where the ( r + 1)th term = x r r 2 0 1 n We can extend this rule to give:
n ( r + 1)th term = a n − r b r r
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n n n n ( a + b ) n = a n + a n −1b1 + a n − 2 b 2 + … + b n where the 1 2 0 n
We can also write the expansion of (1 + x ) n as: (1 + x ) n = 1 + nx +
n( n − 1) 2 n( n − 1)( n − 2) 3 x + x + … + xn 2! 3!
Arithmetic series
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For an arithmetic progression with first term a, common difference d and n terms: ●●
the kth term = a + ( k − 1)d
the last term = l = a + ( n − 1)d n n ( a + l ) = [2 a + ( n − 1)d ]. ●● the sum of the terms = Sn = 2 2 ●●
Geometric series
For a geometric progression with first term a, common ratio r and n terms:
●●
the kth term = ar k −1
●●
the last term = ar n −1
●●
sum of the terms = Sn =
a (1 − r n ) a ( r n − 1) = . 1− r r −1
The condition for a geometric series to converge is, −1 < r < 1 . When a geometric series converges, S∞ =
a . 1− r
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END-OF-CHAPTER REVIEW EXERCISE 6 5
3 1 Find the coefficient of x 2 in the expansion of 2 x + 2 . x
[3]
2 In the expansion of ( a + 2 x )6, the coefficient of x is equal to the coefficient of x 2. Find the value of the constant a.
5 In the expansion of (2 + ax )7 , where a is a constant, the coefficient of x is −2240. Find the coefficient of x 2 . 5
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2 6 Find the coefficient of x5 in the expansion of x 3 + 2 . x
[3]
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x 3 In the expansion of 1 − (5 + x )6, the coefficient of x 2 is zero. Find the value of a. a 6 2 . 4 Find the term independent of x in the expansion of 3x − 5x
[3]
[3]
[4]
[4]
5
1 . 7 Find the term independent of x in the expansion of 3x 2 − 2x3 8 i
Find the first 3 terms in the expansion of ( x − 3x 2 )8 in descending powers of x. 15
2 8
ii Find the coefficient of x in the expansion of (1 − x )( x − 3x ) . 9 i
2
ii Given that the coefficient of x in the expansion of (1 − 2 x )(1 + px ) is 204, find the possible values of p. 10 i
[3] [2] [3]
8
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Find the first three terms in the expansion of (1 + px )8 in ascending powers of x.
[4]
[4]
Find the first three terms, in ascending powers of x, in the expansion of:
a (1 + 2 x )5
[2]
b (3 − x )5.
[2]
ii Find the coefficient of x 2 in the expansion of [(1 + 2 x )(3 − x )]5.
11 The first term of an arithmetic progression is 1.75 and the second term is 1.5. The sum of the first n terms is −n . Find the value of the positive integer n.
[3]
[4]
12 The second term of a geometric progression is −1458 and the fifth term is 432. Find: i
the common ratio
[3]
ii the first term
[1]
iii the sum to infinity.
[2]
13 An arithmetic progression has first term a and common difference d . Give that the sum of the first 100 terms is 25 times the sum of the first 20 terms. i
Find d in terms of a.
ii Write down an expression, in terms of a, for the 50th.
[3] [2]
14 The tenth term of an arithmetic progression is 17 and the sum of the first 5 terms is 190. i
Find the first term of the progression and the common difference.
[4]
ii Given that the nth term of the progression is −19, find the value of n.
[2]
Original material © Cambridge University Press 2017
Chapter 6: Series
CROSS-TOPIC REVISION EXERCISE 2
2 3
4
4
Find the highest power of x in the expansion of (5x 4 + 3)8 + (1 − 3x 3 )5 (4x 2 − 5x5 )6 . 6 1 Find the term independent of x in the expansion of 4x − 2 . x 6 2 i Find the first 3 terms in the expansion of 3x − 2 in descending powers of x. x
[2] [3] [3]
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1
6 2 2 ii Hence find the coefficient of x 2 in the expansion of 1 + 3x − . x x
[2]
i Find the first 3 terms when (1 − 2 x )5 is expanded in ascending powers of x.
[3]
ii In the expansion of (3 + ax )(1 − 2 x )5, the coefficient of x 2 is zero. 5
6
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Find the value of a.
[2]
The first term of a geometric progression is 50 and the second term is −40. i Find the fourth term.
[3]
ii Find the sum to infinity.
[2]
The first three terms of a geometric progression are 3k + 14, k + 14 and k respectively. All the terms in the progression are positive. i Find value of k.
ii Find the sum to infinity.
[2]
The sum of the 1st and 2nd terms of a geometric progression is 50 and the sum of the 2nd and 3rd terms is 30. Find the sum to infinity.
[6]
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Cambridge International AS & A Level Mathematics 9709 Paper 11 Q5 November 2016
4
i Show that cos x ≡ 1 − 2 sin2 x + sin 4 x.
[1]
ii Hence or otherwise, solve the equation 8 sin 4 x + cos 4 x = 2 cos2 x for 0° < x < 360°.
[5]
9
[3]
Cambridge International AS & A Level Mathematics 9709 Paper 11 Q6 November 2016
A sector of a circle, radius r cm , has a perimeter of 60 cm.
i Show that the area, A cm 2, of the sector is given by A = 30 r − r 2.
[2]
ii Express 30 r − r 2 in the form a − ( r − b )2, where a and b are constants.
[2]
Given that r can vary:
iii find the value of r at which A is stationary
[1]
iv find this stationary value of A, and state whether it is a maximum or a minimum.
[2]
10
rcm
r cm
x
The diagram shows a metal plate consisting of a rectangle with sides x cm and r cm and two identical sectors of a circle of radius r cm . The perimeter of the plate is 100 cm.
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