AS-Level Maths: Core 2 for Edexcel

AS-Level Maths: Core 2 for Edexcel
C2.3 Sequences and series These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. 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 © Boardworks Ltd 2005

Contents Geometric sequences Geometric sequences Geometric series
The sum to infinity of a geometric series Binomial expansions Examination-style questions Contents 2 of 39 © Boardworks Ltd 2005

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 × 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, ar5, … ar4, The nth term of a geometric sequence with first term a and common ratio r is: un = arn–1 Also: The formula for the nth term is sometimes called a deductive definition, since it defines the value of un directly, while an inductive definition defines each term in relation to a previous term. The inductive definition of a geometric sequence with first term a and common ratio r is: u1 = a, un+1 = run

What is the 7th term of the geometric sequence 8, 12, 18, 27, …?
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 = 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? Explain that to find the common ratio of a geometric series we can divide any two consecutive terms. For the sequence in the first example we can think of each term as being 50% bigger than the last term. For the second problem explain that we can find the values of the first term a and common ratio r by using the information given to write a pair of simultaneous equations. Using the 3rd term: ar2 = 36 Using the 6th term: ar5 = 972

Geometric sequences Dividing these gives: So: r3 = 27 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.

Find the missing terms Challenge students to find the missing terms in each geometric sequence. As an additional task they could also find an expression for the general term of the sequence.

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 . This gives the following sequence: 8, 2, 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 , we have: Explain that for a sequence to converge to 0 it gets closer and closer to 0 without actually reaching it. 6, –3, This sequence also converges to 0.

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: Explain that we write 0 < |r| < 1 as oppose to –1 < r < 1 to exclude the case where r = 0. |0.6| = 0.6 and |–0.6| = 0.6 In general: For a geometric sequence, if 0 < |r| < 1 the sequence will converge.

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: Explain that when we call a number large it can be either large and positive or large and negative. The further the number is from 0, the larger it is. In theory, a geometric sequence could also have a common ratio of 1 or 0. In the first case all the terms would be the same and in the second all the terms after the first would be 0. Sequences where r = 1 and r = 0 are normally excluded from the definition of a geometric sequence, not only because they are trivial, but also because the formulae associated with geometric progressions can break down when used with these values. 4, –4, 4, –4, 4, –4, … This sequence oscillates between two values.

Contents Geometric series Geometric sequences Geometric series

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) = = 242 When n is large, a more systematic approach for calculating the sum of a given number of terms is required.

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)

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: 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 – –, …

Using Σ notation We can write the sum of a geometric series using Σ notation as: Find This is a geometric series with first term 5 and common ratio 2. There are 7 terms in this sequence so: = 635

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 Contents 16 of 39 © Boardworks Ltd 2005

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 more terms are added. For example, the geometric series tends to 2 as the number of terms increases. We can show this diagrammatically as follows: 2 1 1

If we use the formula Sn = with a = 1 and r = we have Define S∞ as the limit of Sn as n tends to infinity. As n So:

In general, the sum of the first n terms of a geometric series is: But if |r| < 1, In this case, we can write the sum to infinity as: For example: Point out that from this formula we can see that when the common ratio is between 0 and 1, the sum to infinity will be more than the first term (because we divide a by a number less than 1). In contrast, when the common ratio is between –1 and 0, the sum to infinity will be less than the first term (because we divide a by a number greater than 1). This is demonstrated by the example. Find the sum to infinity of the geometric series with first term 6 and common ratio –0.2. = 5

The first term of a geometric series is 20 and the sum to infinity is 15. What is the common ratio? Using we have

Contents Binomial expansions Geometric sequences Geometric series

Pascal’s Triangle

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 Coefficients of 1 have been written to draw attention to the pattern in the coefficients. Students should notice immediately that the coefficients form the same pattern of numbers as those given by Pascal’s triangle. They should also notice that for each term the index for a goes down one each time while the index for b goes up one. For each term the sum of the powers of a and b is the same as the power of (a + b). (a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4 (a + b)5 = 1a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5 What patterns do you notice?

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: Tell students that it is the (n + 1)th row that starts with 1, n. This is because the 1st row is 1, the 2nd row is 1, 1 and the 3rd row is 1, 2, 1 and so on. 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

Binomial expansions Expand (2x – y)4 Using
(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 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 x2y2 – 8xy3 + y4 Notice that when the second term in a binomial is negative the signs of the terms in the expansion will alternate. Ask students to explain why the terms in the expansion will always alternate when the signs of the terms in a binomial are different. Suppose we wanted to expand (a + b)20. We could find the 20th row of Pascal’s triangle, but this would take a very long time.

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 Ways of getting a3b Ways of getting a2b2 Ways of getting ab3 Ways of getting b4 aaaa aaab aabb abbb bbbb aaba abab babb abaa abba bbab Explain that when multiplying out the brackets there is only one way to get the term in a4, that is by multiplying the a in the first bracket by each of the a’s in the other brackets. There are four ways to get the term in a3b. That is by choosing a b from one of the four brackets and three a’s from the remaining three brackets. The coefficient of a3b is therefore 4 in the expansion of (a + b)4. Continue the argument for the remaining terms. baaa bbaa bbba baba baab 1 way 4 ways 6 ways 4 ways 1 way

The situation where no b’s (or four a’s) are chosen from any of the four brackets can be written as 4C0 or . This is the same as 4C4 or . The situation where one b (or three a’s) can be chosen from any of the four brackets can be written as: 4C1 or . This is the same as 4C3 or . Tell students that 4C0 is read as ‘4 choose 0’. Explain that it makes no real difference whether we consider choosing a’s or choosing b’s. Choosing no b’s from the four brackets is exactly the same as choosing four a’s from the four brackets (or choosing four b’s or no a’s). 4C0 is therefore equivalent to 4C4 and 4C1 is equivalent to 4C3. Using this argument we can see the reason for the symmetry of the binomial coefficients. The situation where two b‘s (or two a’s) can be chosen from any of the four brackets can be written as 4C2 or

The fifth row of Pascal’s triangle can be written as: This corresponds to the values 1 4 6 The expansion of (a + b)4 can therefore be written as: Remind students that it is the 5th row of Pascal’s triangle that starts 1,4 because the 1st row is 1, the 2nd row is 1, 1 and the 3rd row is 1, 2, 1. Or: (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

The number of ways to choose r objects from a group of n objects is written as nCr and is given by n! is read as ‘n factorial’ and is the product of all the natural numbers from 1 to n. In general: The proof of this formula is given by the theory of combinations, which is outside the scope of this presentation. It may be investigated as a separate activity if desired. n! = n × (n –1) × (n – 2) × (n – 3) … × 2 × 1 n can also be 0 and by definition 0! = 1.

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 = 20! = 20 × 19 × 18 × … × 2 × 1 = Fortunately, when we use the formula Advise students to find the n! key on their calculators. They can then experiment in finding the value of n! for different values of n. Discuss the calculation of 4C2 and remind students that this is the coefficient of a2b2 in the expansion of (a + b)n. to calculate binomial coefficients, many of the numbers cancel out. For example, for 4C2 we have 2 = 6

Here are some more examples: 56 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 36 Students can check these by using the nCr button on their calculators. Point out that (85) will give the same value as (83) and so 56 is also the coefficient of a5b3 in the expansion of (a + b)8. This is because the number of ways of choosing 3 a’s from 8 brackets is the same as the number of ways of choosing 3 b’s from 8 brackets. Similarly, (92) will give the same value as (97) and so 36 is also the coefficient of a2b7 in the expansion of (a + b)9. 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.

The effect of this cancelling gives an alternative form for nCr. In general, the expansion of (a + b)n can be written as: A special case is the expansion of (1 + x)n

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 = 120(–8a7b3) = –960a7b3 So the coefficient of a7b3 in the expansion of (a – 2b)10 is –960.

Use the binomial theorem to write down the first four terms in the expansion of (1 + x)7 in ascending powers of x. 3 If necessary, explain that to arrange the terms in ascending powers of x means starting with the lowest power of x and ending with the highest. How could we use this expansion to find an approximate value for 1.17?

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 × 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. If the expansion is completed in full for ( )7 we get the exact value of 1.17 as = The given approximation is therefore accurate to two decimal places. 1.112 ≈ ≈ 1.945

Examination-style questions
Geometric sequences Geometric series The sum to infinity of a geometric series Binomial expansions Examination-style questions Contents 36 of 39 © Boardworks Ltd 2005

Examination-style question 1
The 2nd term of a geometric series is 40 and the 5th term is Find the value of the first term and the common ratio. Calculate the sum of the first 10 terms of the series. Calculate the sum to infinity. a) Using the 2nd term: ar = 40 Using the 5th term: ar4 = 20.48 Dividing these gives: So: r3 = 0.152 r = 0.8

Substituting this into ar = 40 gives: a × 0.8 = 40 a = 50 b) Using with n = 10, a = 50 and r = 0.8 gives: = (to 2 d.p.) c) Using with a = 50 and r = 0.8 gives: = 250

Write down the first four terms in the expansion of (1 + ax)13 in ascending powers of x, where a > 0. 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 + ax)13 = = ax + 78a2x a3x3 + … b) 13a = –b 1 78a2 = 12b 2 78a2 = 12 × –13a Substituting into : 1 2 78a = –156 a = –2 b = 26

AS-Level Maths: Core 2 for Edexcel

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