#### Transcript C2.3 Sequences and series

```AS-Level Maths:
Core 2
for Edexcel
C2.3 Sequences
and series
This icon indicates the slide contains activities created in Flash. These activities are not editable.
For more detailed instructions, see the Getting Started presentation.
1 of 39
Geometric sequences
Contents
Geometric sequences
Geometric series
The sum to infinity of a geometric series
Binomial expansions
Examination-style questions
2 of 39
Geometric sequences
In a geometric sequence (or geometric progression) each
term is produced by multiplying the previous term by a
constant value called the common ratio.
For example, the sequence
3, 6, 12, 24, 48, 96, …
is a geometric sequence that starts with 3 and has a common
ratio of 2.
We could write this sequence as
3, 3 × 2, 3 × 2 × 2, 3 × 2 × 2 × 2, 3 × 2 × 2 × 2 × 2, …
or
3,
3 of 39
3 × 2,
3 × 22,
3 × 23,
3 × 24,
3 × 25, …
Geometric sequences
If we call the first term of a geometric sequence a and the
common ratio r we can write a general geometric sequence
as:
a,
ar,
ar2,
ar3,
ar4,
ar5, …
The nth term of a geometric sequence with first term
a and common ratio r is:
un = arn–1
Also:
The inductive definition of a geometric sequence
with first term a and common ratio r is:
u1 = a, un+1 = run
4 of 39
Geometric sequences
What is the 7th term of the geometric
sequence 8, 12, 18, 27, …?
This is a geometric sequence with first term a = 7 and common
ratio r = 12 ÷ 8 = 1.5.
The nth term is given by arn–1 so the 7th term is:
u7 = 8 × 1.56
= 91.125
The 3rd term in a geometric sequence is 36 and the 6th term is
972. What is the value of the 1st term and the common ratio?
Using the 3rd term:
Using the 6th term:
5 of 39
ar2 = 36
ar5 = 972
Geometric sequences
Dividing these gives:
ar 5 972
=
2
ar
36
r3 = 27
So:
r=3
Substituting this into ar2 = 36 gives:
a × 32 = 36
9a = 36
a=4
So the first term of the sequence is 4 and the common ratio is 3.
The general term of this sequence is un = 4 × (3)n–1.
6 of 39
Find the missing terms
7 of 39
Convergent and divergent sequences
Geometric sequences either converge or diverge depending
on the value of the common ratio r.
Suppose the first value of a geometric sequence is 8 and the
common ratio is 41 . This gives the following sequence:
8,
2,
1
2
,
1
8
,
1
32
,
1
128
, ...
This sequence converges to 0.
If the common ratio is negative the signs of the terms will
alternate. For example, if the first value of a geometric
sequence is 6 and the common ratio is  21 , we have:
6,
–3,
3
2
,
 34 ,
3
8
,
 163 , ...
This sequence also converges to 0.
8 of 39
Convergent and divergent sequences
In general, if the common ratio r of a geometric sequence is
between 0 and 1 or between –1 and 0, the terms of the
sequence will converge to 0.
We use |r| to represent the modulus of r, mod r.
This is the numerical value of r, regardless of whether it is
positive or negative.
So, for example:
|0.6| = 0.6
and
|–0.6| = 0.6
In general:
For a geometric sequence, if 0 < |r| < 1
the sequence will converge.
9 of 39
Convergent and divergent sequences
For a geometric sequence, if |r| > 1 the
sequence will diverge.
This means that the terms will get larger and larger without limit.
For example, if the first value of a geometric sequence is 0.2
and the common ratio is –5, the sequence will be:
0.2,
–1,
5,
–25,
125,
–625, …
This sequence is divergent.
The only geometric sequences that neither converge nor
diverge are those where the common ratio is –1. For example:
4,
–4,
4,
–4,
4,
–4, …
This sequence oscillates between two values.
10 of 39
Contents
Geometric series
Geometric sequences
Geometric series
The sum to infinity of a geometric series
Binomial expansions
Examination-style questions
11 of 39
Geometric series
The sum of all the terms of a geometric sequence is called a
geometric series.
We can write the sum of the first n terms of a geometric series
as:
Sn = a + ar + ar2 + ar3 + … + arn–1
For example, the sum of the first 5 terms of the geometric
series with first term 2 and common ratio 3 is:
S4 = 2 + (2 × 3) + (2 × 32) + (2 × 33) + (2 × 34)
= 2 + 6 + 18 + 54 + 162
= 242
When n is large, a more systematic approach for calculating the
sum of a given number of terms is required.
12 of 39
The sum of a geometric series
Start by writing the sum of the first n terms of a general
geometric series with first term a and common ratio r as:
Sn = a + ar + ar2 + ar3 + … + arn–1
Multiplying both sides by r gives:
rSn = ar + ar2 + ar3 + … + arn–1 + arn
Now if we subtract the first equation from the second we have:
rSn – Sn= arn – a
Sn(r – 1) = a(rn – 1)
a( r n  1)
Sn =
r 1
13 of 39
The sum of a geometric series
If we multiply the numerator and the denominator by –1 we
can also write the sum of the first n terms as:
a(1  r n )
Sn =
1 r
This form is more useful when |r| < 1 since it avoids the use of
negative numbers.
For example:
Find the sum of the first 8 terms of the
geometric series that starts 4 – 3 + 2 41 –, …
14 of 39
Using Σ notation
We can write the sum of a geometric series using Σ notation as:
n
n
a
(
r

1)
a
(1

r
)
i 1
ar =
=

r 1
1 r
i =1
n
7
Find
i 1
5(2)

i =1
7
i 1
6
5(2)
=
5
+10
+
20
+...
+
5(2)

i =1
This is a geometric series with first term 5 and common ratio 2.
There are 7 terms in this sequence so:
7
5(2
 1)
i 1
5(2) =

2 1
i =1
= 635
7
15 of 39
Contents
The sum to infinity of a geometric series
Geometric sequences
Geometric series
The sum to infinity of a geometric series
Binomial expansions
Examination-style questions
16 of 39
The sum to infinity of a geometric series
When the common ratio of a geometric series is between –1
and 1, the sum of the series will tend to a particular value as
For example, the geometric series
1 + 21 + 41 + 81 +
1
16
+ ...
tends to 2 as the number of terms increases.
We can show this diagrammatically as follows:
2
1
2
1
1
1
16
1
8
17 of 39
1
4
1
2
The sum to infinity of a geometric series
a(1  r n )
If we use the formula Sn =
with a = 1 and r = 21 we have
1 r
1(1  ( 21 )n )
Sn =
1  21
=
1  ( 21 )n
1
2
= 2  ( 21 )n1
=2
1
As n  , n1  0.
2
So:
18 of 39
1
2n1
S = 2
where S = lim Sn
n
The sum to infinity of a geometric series
In general, the sum of the first n terms of a geometric series is:
a(1  r n )
Sn =
1 r
But if |r| < 1, r n  0 as n  
In this case, we can write the sum to infinity as:
a
S =
1 r
For example:
Find the sum to infinity of the geometric series
with first term 6 and common ratio –0.2.
6
S =
=5
1+ 0.2
19 of 39
The sum to infinity of a geometric series
The first term of a geometric series is 20 and the
sum to infinity is 15. What is the common ratio?
a
Using S =
we have
1 r
20
15 =
1 r
15(1 r ) = 20
20
1 r =
15
20
r = 1
15
1
r=
3
20 of 39
Contents
Binomial expansions
Geometric sequences
Geometric series
The sum to infinity of a geometric series
Binomial expansions
Examination-style questions
21 of 39
Pascal’s Triangle
22 of 39
Binomial expansions
An expression containing two terms, for example a + b, is called
a binomial expression.
When we find powers of binomial expressions an interesting
pattern emerges.
(a + b)0 = 1
(a + b)1 = 1a + 1b
(a + b)2 = 1a2 + 2ab + 1b2
(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
(a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4
(a + b)5 = 1a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5
What patterns do you notice?
23 of 39
Binomial expansions
In general, in the expansion of (a + b)n:
The coefficients are given by the (n + 1)th row of Pascal’s
triangle.
The sum of the powers of a and b is n for each term.
Altogether, there are n + 1 terms in the expansion.
As long as n is relatively small, we can expand a given binomial
directly by comparing it to the equivalent expansion of (a + b)n.
For example:
Expand (x + 1)5.
Using (a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5
and replacing a with x and b with 1 gives:
(x + 1)5 = x5 + 5x4 + 10x3 + 10x2 + 5x + 1
24 of 39
Binomial expansions
Expand (2x – y)4.
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Using
and replacing a with 2x and b with –y gives:
(2x – y)4 = (2x)4 + 4(2x)3(–y) + 6(2x)2(–y)2 + 4(2x)(–y)3 + (–y)4
= 16x4 – 32x3y
+
24x2y2
–
8xy3
+ y4
Notice that when the second term in a binomial is negative the
signs of the terms in the expansion will alternate.
Suppose we wanted to expand (a + b)20.
We could find the 21st row of Pascal’s triangle, but this would
take a very long time.
25 of 39
Finding binomial coefficients
When n is large we can find the binomial coefficients using
combinations theory.
Let’s look more closely at the expansion of
(a + b)4 = (a + b)(a + b)(a + b)(a + b)
Ways of
getting a4
aaaa
1 way
26 of 39
Ways of
Ways of
Ways of
getting a3b getting a2b2 getting ab3
aaab
aabb
abbb
aaba
abab
babb
abaa
abba
bbab
baaa
bbaa
bbba
baba
baab
4 ways
6 ways
4 ways
Ways of
getting b4
bbbb
1 way
Finding binomial coefficients
The situation where no b’s (or four a’s) are chosen from any of
the four brackets can be written as
 4
 4
4
This is the same as C4 or   .
or   .
0
 4
The situation where one b (or three a’s) can be chosen from
any of the four brackets can be written as:
4C
0
4
4


4C or
 1  . This is the same as 4C3 or  3  .
1
 
 
The situation where two b’s (or two a’s) can be chosen from
any of the four brackets can be written as
4C
2
27 of 39
 4
or   .
 2
Finding binomial coefficients
The fifth row of Pascal’s triangle can be written as:
 4
0
 
 4
 1
 
 4
 2
 
 4
3
 
 4
 4
 
4
1
This corresponds to the values
1
4
6
The expansion of (a + b)4 can therefore be written as:
 4 4  4 3  4 2 2  4 3  4 4
(a  b)    a    a b    a b    ab    b
0
 1
 2
3
 4
4
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4
Or:
28 of 39
Finding binomial coefficients
The number of ways to choose r objects from a group of
n objects is written as nCr and is given by
n
n!
 r  = r! n  r !
 
 
n! is read as ‘n factorial’ and is the product of all the natural
numbers from 1 to n.
In general:
n! = n × (n –1) × (n – 2) × (n – 3) … × 2 × 1
n can also be 0 and by definition 0! = 1.
29 of 39
Finding binomial coefficients
The value of n! gets large very quickly as the value of n
increases. For example:
5! = 5 × 4 × 3 × 2 × 1 = 120
12! = 12 × 11 × 10 × … × 2 × 1 = 479 001 600
20! = 20 × 19 × 18 × … × 2 × 1 = 2 432 902 008 176 640 000
Fortunately, when we use the formula
n
n!
 r  = r! n  r !
 
 
to calculate binomial coefficients, many of the numbers cancel
out. For example, for 4C2 we have
2
4 × 3 × 2×1
 4
4!
4×3
=
=
=
=6
 2  2! 2! (2×1)×(2×1)
2 ×1
 
30 of 39
Finding binomial coefficients
Here are some more examples:
8 × 7 × 6 × 5 × 4 × 3 × 2×1
8
8!
8×7×6
=
=
=
= 56
 3  3! 5! (3 × 2×1)×(5 × 4 × 3 × 2×1)
3 × 2 ×1
 
This value corresponds to the number of ways of choosing 3 a’s
from the 8 brackets in the expansion of (a + b)8.
56 is therefore the coefficient of a3b5 in the expansion of (a + b)8.
4
9× 8× 7× 6× 5× 4× 3× 2×1
9
9!
9×8
=
=
=
= 36
 7  7! 2! (7× 6× 5× 4× 3× 2×1)×(2×1)
2 ×1
 
This value corresponds to the number of ways of choosing 7 a’s
from the 9 brackets in the expansion of (a + b)9.
36 is therefore the coefficient of a7b2 in the expansion of (a + b)9.
31 of 39
Finding binomial coefficients
The effect of this cancelling gives an alternative form for nCr.
 n  n ×( n  1)×( n  2)×...×( n  r +1)
r =
r!
 
In general, the expansion of (a + b)n can be written as:
 n  n  n  n 1  n  n 2 2
n n
(a + b) =   a +   a b +   a b + ...+   b
0
 1
 2
n
n
A special case is the expansion of (1 + x)n
 n n
n 2 n 3
(1 + x ) =   +   x +   x +   x +... + x n
 0   1
 2
3
n( n  1) 2 n( n  1)( n  2) 3
= 1+ nx +
x +
x +... + x n
2!
3!
n
32 of 39
Using the binomial theorem
This method of finding the binomial coefficients is called the
binomial theorem.
Find the coefficient of a7b3 in the
expansion of (a – 2b)10.
The term in a7b3 is of the form:
3
4
 10  7
10 × 9 × 8 7
3
3
a
(

2
b
)
=
a
(

8
b
)
3
1× 2× 3
 
= 120(–8a7b3)
= –960a7b3
So the coefficient of a7b3 in the expansion of (a – 2b)10 is –960.
33 of 39
Using the binomial theorem
Use the binomial theorem to write down the first four terms
in the expansion of (1 + x)7 in ascending powers of x.
7 7
7 2 7 3
(1+ x ) =   +   x +   x +   x +...
 0   1
 2
3
7
3
= 1+ 7 x +
7×6 2 7×6×5 3
x +
x +...
2×1
3 × 2×1
= 1+ 7 x + 21x2 + 35 x3 +...
How could we use this expansion to find an
approximate value for 1.17?
34 of 39
Using the binomial theorem
To find an approximate value for 1.17 we can let x = 0.1 in the
expansion
(1 + x)7 = 1 + 7x + 21x2 + 35x3 + …
This gives us
1.17 ≈ 1 + 7 × 0.1 + 21 × 0.12 + 35 × 0.13
As 0.1 is raised to ever higher powers it becomes much smaller
and so less significant.
We can therefore leave out higher powers of x and still have a
reasonable approximation.
1.17 ≈ 1 + 0.7 + 0.21 + 0.035
≈ 1.945
35 of 39
Contents
Examination-style questions
Geometric sequences
Geometric series
The sum to infinity of a geometric series
Binomial expansions
Examination-style questions
36 of 39
Examination-style question 1
The 2nd term of a geometric series is 40 and the 5th term is
20.48.
a) Find the value of the first term and the common ratio.
b) Calculate the sum of the first 10 terms of the series.
c) Calculate the sum to infinity.
a) Using the 2nd term:
ar = 40
Using the 5th term:
ar4 = 20.48
Dividing these gives:
ar 4 20.48
=
ar
40
r3 = 0.512
So:
r = 0.8
37 of 39
Examination-style question 1
Substituting this into ar = 40 gives:
a × 0.8 = 40
a = 50
a( r n  1)
b) Using Sn =
with n = 10, a = 50 and r = 0.8 gives:
r 1
50(0.810  1)
S10 =
0.8  1
= 223.16 (to 2 d.p.)
a
c) Using S =
with a = 50 and r = 0.8 gives:
1 r
50
S =
1  0.8
= 250
38 of 39
Examination-style question 2
a) Write down the first four terms in the expansion of (1 + ax)13
in ascending powers of x, where a > 0.
b) Given that in the expansion of (1 + ax)13 the coefficient of x is
–b and the coefficient of x2 is 12b, find the value of a and b.
a)
(1 +
13 ×12
13 ×12×11
2
1+13ax +
( ax ) +
( ax )3 +...
2
3×2
= 1 + 13ax + 78a2x2 + 286a3x3 + …
ax)13 =
13a = –b 1
78a2 = 12b 2
78a2 = 12 × –13a
Substituting 1 into 2 : 78a = –156
a = –2
b = 26
b)
39 of 39