DP Maths AI SL 1.9 The Binomial Theorem

DP Maths AI SL 1.9 The Binomial Theorem - Full notes

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DP Maths AI SL | 1.9 The Binomial Theorem

DP MATHEMATICS AI · TOPIC 1: NUMBER AND ALGEBRA

1.9

The Binomial Theorem

Full notes for Applications and Interpretation SL

Contents

1	Expanding brackets and spotting patterns
2	Pascal's triangle
3	Factorials
4	Binomial coefficients and counting
5	Properties of binomial coefficients
6	The binomial theorem
7	Expansions with negative and fractional coefficients
8	Finding a specific term
9	The term independent of x
10	Finding unknown constants from coefficients
11	Products of expansions
12	Approximations using expansions
13	Counting applications: selections and paths
14	Link to probability: the binomial distribution
15	Patterns in Pascal's triangle
16	Approximations in finance and science
17	Using technology
18	Mixed review: worked problems
19	Investigation: coin sequences and Pascal's triangle
20	Exam-style questions with markscheme
21	Common misconceptions and links
22	Check your understanding
23	Summary and topic map
24	Answers to practice exercises

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and work through every worked example before attempting the practice exercise.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
expand (a + b)n using Pascal's triangle
evaluate factorials and binomial coefficients nCr
use the binomial theorem and the general term
find a specific term, a coefficient or the term independent of x
find unknown constants from given coefficients
use expansions to approximate
apply binomial coefficients to counting and probability

SL booklet: this booklet contains the content for Applications and Interpretation SL. Sections marked for HL in the HL edition are not included.

Guiding questions

How can powers of a sum be expanded efficiently? Why do the same numbers appear in counting, probability and algebra?

How this topic is examined

Binomial expansions are tested on Paper 1 (exact expansions, coefficients, the constant term) and support the binomial distribution in Paper 2. Questions are usually short, but sign and power errors are very common.

Write the general term, collect the powers of x, and only then substitute numbers. Show each coefficient calculation.

Mathematics: Applications and Interpretation places strong emphasis on technology, modelling and interpretation. Most papers allow a GDC, so you are expected to use it efficiently, record the method you used, and interpret every answer in context.

Command term	What the examiner wants
Write down	an answer with little or no working; it should be obtainable by inspection
Find / Calculate / Determine	obtain the answer showing relevant working, including what you entered into a GDC
Show that	obtain the given result; every step must be shown and the result must not be assumed
Estimate	obtain an approximate value
Sketch	a diagram showing the general shape and key features, labelled
Interpret / Comment	explain the meaning of a result in context, or give a judgement
Justify / Explain	give valid reasons or evidence
Hence	use the previous result

How to use these notes

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or explanation of a key result
Key point	a result or idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
GDC tip	how to use your graphic display calculator efficiently
In the real world	where the mathematics is used outside the classroom
Worked example	a full solution with IB-style marks (M1 method, A1 answer, R1 reasoning, AG answer given), graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet

1 Expanding brackets and spotting patterns

Multiply out and look.

Think about it

Expand (a + b)2 and (a + b)3 by hand. What patterns do you notice in the powers and in the coefficients?

A binomial is an expression with two terms, such as a + b, x − 3 or 2x + 5. Expanding powers of a binomial by repeated multiplication becomes slow very quickly. Looking at the first few powers reveals a pattern:

Power	Expansion
(a + b)1	a + b
(a + b)2	a2 + 2ab + b2
(a + b)3	a3 + 3a2b + 3ab2 + b3
(a + b)4	a4 + 4a3b + 6a2b2 + 4ab3 + b4

In (a + b)n: there are n + 1 terms; the powers of a decrease from n to 0; the powers of b increase from 0 to n; the powers in every term add to n; and the coefficients are symmetric.

WORKED EXAMPLE 1 Expanding by multiplication (basic)

Expand (x + 2)3 by multiplying out.

Solution

(x + 2)2 = x2 + 4x + 4A1

(x2 + 4x + 4)(x + 2) = x3 + 6x2 + 12x + 8A1

Practice exercise 1

1. Expand (x − 3)2.

2. Expand (2x + 1)3 by multiplying out.

3. How many terms are in (a + b)12?

2 Pascal's triangle

Each entry is the sum of the two above.

The coefficients of (a + b)n form Pascal's triangle. Each row begins and ends with 1, and every inner number is the sum of the two numbers directly above it. Row n (starting with row 0) gives the coefficients of (a + b)n.

n	Row of Pascal's triangle
0	1
1	1 1
2	1 2 1
3	1 3 3 1
4	1 4 6 4 1
5	1 5 10 10 5 1
6	1 6 15 20 15 6 1
7	1 7 21 35 35 21 7 1
8	1 8 28 56 70 56 28 8 1

Why it works

(a + b)n+1 = (a + b)(a + b)n. Each term of the new expansion comes from multiplying a term of (a + b)n by a or by b. The coefficient of an+1−rbr therefore collects two contributions: one from the term with br (times a) and one from the term with br−1 (times b). Those are the two numbers above it in the triangle.

WORKED EXAMPLE 2 Using the triangle (standard)

Expand (x + 3)4.

Solution

Coefficients 1, 4, 6, 4, 1M1

x4 + 4x3(3) + 6x2(9) + 4x(27) + 81M1

= x4 + 12x3 + 54x2 + 108x + 81A1

Practice exercise 2

1. Write down row 9 of Pascal's triangle.

2. Expand (y + 2)5.

3. Find the sum of the numbers in row 10.

3 Factorials

n! = n × (n − 1) × ... × 1.

The factorial of a non-negative integer n is the product of all positive integers up to n, with 0! = 1. Factorials count arrangements: 5 different books can be arranged in 5! = 120 orders. They grow extremely fast: 10! = 3 628 800 and 20! ≈ 2.4 × 1018.

Many factorial expressions simplify by cancelling: 10!/8! = 10 × 9 = 90.

WORKED EXAMPLE 3 Simplifying factorials (standard)

Simplify (a) 12!/10!; (b) (n + 1)!/(n − 1)!.

Solution

(a) 12 × 11 = 132A1

(b) (n + 1)nA1

Practice exercise 3

1. Evaluate 7!.

2. Simplify 20!/(18! 2!).

3. In how many ways can 6 runners finish a race?

4 Binomial coefficients and counting

Choosing r from n.

The binomial coefficient nCr counts the number of ways of choosing r objects from n when order does not matter. It is also the entry in row n, position r (starting from r = 0) of Pascal's triangle. For example, 5C2 = 10: there are 10 ways to choose 2 people from 5, and the coefficient of a3b2 in (a + b)5 is 10.

Why it works

In expanding (a + b)n, each term arises by choosing a or b from each of the n brackets. The term an−rbr arises whenever b is chosen from exactly r brackets, and there are nCr ways of choosing those brackets.

WORKED EXAMPLE 4 Evaluating nCr (basic)

Evaluate 8C3 without a calculator.

Solution

8!/(3! 5!) = (8 × 7 × 6)/(3 × 2 × 1)M1

= 56A1

WORKED EXAMPLE 5 A counting problem (standard)

A committee of 4 is chosen from 9 people. How many committees are possible?

Solution

9C4M1

= 126A1

Practice exercise 4

1. Evaluate 10C4.

2. In how many ways can 3 toppings be chosen from 12?

3. Solve nC2 = 36.

5 Properties of binomial coefficients

Symmetry and Pascal's rule.

Symmetry: nCr = nC(n − r): choosing r to include is the same as choosing n − r to exclude. Edges: nC0 = nCn = 1, nC1 = n. Pascal's rule: nCr + nC(r + 1) = (n + 1)C(r + 1). Row sum: nC0 + nC1 + ... + nCn = 2n (substitute a = b = 1 into (a + b)n).

Practice exercise 5

1. Show that 7C3 = 7C4.

2. Verify Pascal's rule for 6C2 + 6C3.

3. Find nC0 + nC1 + ... + nCn for n = 7.

6 The binomial theorem

The general formula.

The general term (the (r + 1)th term) is nCr an−r br. Use the theorem when Pascal's triangle would need too many rows.

WORKED EXAMPLE 6 Using the theorem (standard)

Find the first four terms of (1 + 2x)9 in ascending powers of x.

Solution

1 + 9C1(2x) + 9C2(2x)2 + 9C3(2x)3M1

= 1 + 18x + 36(4x2) + 84(8x3)A1

= 1 + 18x + 144x2 + 672x3A1

Practice exercise 6

1. Find the first three terms of (2 + x)7.

2. Write down the general term of (3x + 5)12.

7 Expansions with negative and fractional coefficients

Keep the sign with b.

When the second term is negative, write the binomial as a + (−b): the powers of −b alternate in sign. When a term has a coefficient, raise the whole term to the power: (3x)2 = 9x2, not 3x2. Fractions such as x/2 or 1/x are handled the same way.

WORKED EXAMPLE 7 Negative term (standard)

Expand (2x − 3)4.

Solution

Coefficients 1, 4, 6, 4, 1 with a = 2x, b = −3M1

16x4 + 4(8x3)(−3) + 6(4x2)(9) + 4(2x)(−27) + 81M1

= 16x4 − 96x3 + 216x2 − 216x + 81A1

Common mistake

Writing (2x)4 = 2x4, or losing the sign of the negative term in odd powers.

Practice exercise 7

1. Expand (1 − x)5.

2. Expand (x/2 + 4)3.

3. Expand (x − 1/x)4 and simplify.

8 Finding a specific term

Solve for r.

To find the term in xk: write the general term nCr an−rbr, collect all powers of x into xf(r), solve f(r) = k for r, and evaluate the coefficient. Remember that the term with br is the (r + 1)th term.

WORKED EXAMPLE 8 A specific coefficient (standard)

Find the coefficient of x3 in (4 − x)8.

Solution

General term 8Cr 48−r(−x)r; r = 3M1

8C3 × 45 × (−1)3 = 56 × 1024 × (−1)A1

= −57 344A1

WORKED EXAMPLE 9 Powers of x in both terms (exam level)

Find the coefficient of x7 in (2x2 + 1/x)8.

Solution

8Cr (2x2)8−r(x−1)r = 8Cr 28−r x16−3rM1

16 − 3r = 7, r = 3A1

56 × 25 = 1792A1

Practice exercise 8

1. Find the coefficient of x4 in (1 + 3x)6.

2. Find the coefficient of x2 in (x − 2)7.

3. Find the 5th term of (2 + x)10.

9 The term independent of x

Set the power of x to zero.

WORKED EXAMPLE 10 Constant term (exam level)

Find the term independent of x in (x + 3/x2)9.

Solution

9Cr x9−r(3x−2)r = 9Cr 3r x9−3rM1

9 − 3r = 0, r = 3A1

84 × 27 = 2268A1

Practice exercise 9

1. Find the constant term in (x + 1/x)6.

2. Find the constant term in (2x − 1/x)8.

3. Show that (x + 1/x)5 has no constant term.

10 Finding unknown constants from coefficients

Form an equation.

WORKED EXAMPLE 11 Unknown constant (exam level)

The coefficient of x2 in (1 + kx)6 is 135. Find the possible values of k.

Solution

6C2 k2 = 135M1

15k2 = 135, k2 = 9A1

k = ±3A1

WORKED EXAMPLE 12 Two conditions (exam level)

In (1 + ax)n, the coefficient of x is 12 and the coefficient of x2 is 60. Find a and n.

Solution

na = 12 and n(n − 1)a2/2 = 60M1

Divide the second equation by the first: (n − 1)a/2 = 5, so (n − 1)a = 10; with na = 12, a = 2M1 A1

a = 2, n = 6A1

Practice exercise 10

1. The coefficient of x3 in (2 + px)5 is 720. Find p.

2. The constant term of (x2 + k/x)6 is 240. Find k (k > 0).

11 Products of expansions

Combine contributions.

WORKED EXAMPLE 13 Coefficient in a product (exam level)

Find the coefficient of x2 in (3 − x)(1 + 2x)5.

Solution

(1 + 2x)5 = 1 + 10x + 40x2 + ...A1

x2: 3 × 40 + (−x)(10x) → 120 − 10M1

= 110A1

Practice exercise 11

1. Find the coefficient of x3 in (1 + x)2(1 − x)4.

2. Find the coefficient of x in (2 + x)(1 − 3x)6.

12 Approximations using expansions

Small x, few terms.

When x is small, the terms with high powers of x are tiny, so the first few terms of an expansion give an accurate approximation. This is useful for estimating powers of numbers close to 1, such as 1.0210 or 0.986.

WORKED EXAMPLE 14 Approximating a power (standard)

Use the first three terms of (1 + x)10 to estimate 1.0210.

Solution

1 + 10x + 45x2 with x = 0.02M1

1 + 0.2 + 0.018 = 1.218A1

Actual 1.2190: error about 0.08%R1

Practice exercise 12

1. Use three terms of (1 − x)8 to estimate 0.998.

2. Estimate 2.015 using (2 + x)5.

13 Counting applications: selections and paths

nCr in counting problems.

nCr counts selections when order does not matter: committees, hands of cards, choices of questions. It also counts paths on a grid: to travel from one corner of an m × n grid to the opposite corner using only right and up steps, you choose which m of the (m + n) steps go right: (m + n)Cm paths.

WORKED EXAMPLE 15 Grid paths (standard)

How many shortest paths are there from (0, 0) to (5, 3) moving only right or up?

Solution

8 steps, choose 5 to be rightM1

8C5 = 56A1

WORKED EXAMPLE 16 Selection with a condition (exam level)

A team of 5 is chosen from 7 girls and 6 boys, with exactly 3 girls. How many teams?

Solution

7C3 × 6C2M1

35 × 15 = 525A1

Practice exercise 13

1. How many 5-card hands can be dealt from 52 cards?

2. How many shortest paths from (0, 0) to (6, 6)?

14 Link to probability: the binomial distribution

Same coefficients, new meaning.

If an experiment with success probability p is repeated n times independently, the probability of exactly r successes is nCr pr(1 − p)n−r. These are exactly the terms of the expansion of (q + p)n, where q = 1 − p, which is why the probabilities sum to (q + p)n = 1. The binomial distribution is studied in its own booklet.

WORKED EXAMPLE 17 Coin tosses (standard)

A fair coin is tossed 6 times. Find P(exactly 4 heads).

Solution

6C4 (0.5)6M1

= 15/64 = 0.234A1

Practice exercise 14

1. A die is rolled 5 times. Find P(exactly two sixes).

2. Explain why the probabilities in a binomial distribution add to 1.

15 Patterns in Pascal's triangle

Hidden structure.

Pascal's triangle contains many patterns: the row sums are powers of 2; the diagonals contain the natural numbers, the triangular numbers and the tetrahedral numbers; shallow diagonals add up to the Fibonacci numbers; and colouring the odd entries produces the Sierpinski triangle fractal.

Figure 1. Odd entries (dark) in the first 16 rows of Pascal's triangle form a Sierpinski pattern.

Practice exercise 15

1. Find the sum of the entries in row 6.

2. Write down the third diagonal (1, 3, 6, 10, ...) and name the sequence.

16 Approximations in finance and science

When x is a small rate.

For a small growth rate r, (1 + r)n ≈ 1 + nr + n(n − 1)r2/2. The first-order approximation 1 + nr is the simple-interest value; the second term shows the extra 'interest on interest' from compounding.

WORKED EXAMPLE 18 Compound versus simple (standard)

Estimate the value of $1000 after 5 years at 2% compound interest using three terms, and compare with simple interest.

Solution

1000(1 + 5(0.02) + 10(0.0004)) = $1104M1 A1

Simple interest: $1100; exact compound $1104.08R1

Practice exercise 16

1. Estimate 1.034 using three terms.

2. Explain why the approximation becomes worse for large n or r.

17 Using technology

nCr and expansions on the GDC.

Use the nCr function (probability menu) to evaluate binomial coefficients, and a table of y = nCr x (with x as r) to list a whole row of Pascal's triangle. Some calculators expand polynomials symbolically; on Paper 2 this is acceptable, but on Paper 1 you must show the general term.

GDC tip

To list row 10, enter y = 10 nCr x in the function editor and view the table from x = 0 to 10.

Practice exercise 17

1. Use technology to find 25C12.

2. Find the largest coefficient in the expansion of (1 + x)15.

18 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 19 Mixed problem 1 (exam level)

Find the coefficient of x4 in (x2 − 2/x)8.

Solution

8Cr (x2)8−r(−2)rx−r = 8Cr(−2)rx16−3rM1

16 − 3r = 4, r = 4A1

70 × 16 = 1120A1

WORKED EXAMPLE 20 Mixed problem 2 (exam level)

The first three terms of (1 + ax)n are 1 + 20x + 180x2. Find a and n.

Solution

na = 20; n(n − 1)a2/2 = 180M1

(n − 1)a = 18, so a = 2A1

a = 2, n = 10A1

Practice exercise 18

1. Find the coefficient of x5 in (1 + x)8 + (1 − x)8.

2. Find the term in x3 in (3x − 1)6.

19 Investigation: coin sequences and Pascal's triangle

A Paper 3 style exploration.

1. List all sequences of 4 coin tosses with exactly 2 heads. How many are there? 2. Explain why the number of sequences of n tosses with r heads is nCr. 3. Show that the total number of sequences of n tosses is 2n, and connect this to the row sums of Pascal's triangle. 4. For 10 tosses, find the most likely number of heads and its probability. 5. How does the probability of exactly half heads change as n increases (n = 10, 20, 50, 100)?

Answers. 1. 6. 2. Choose which r of the n positions are heads. 3. Each toss has 2 outcomes; the row sum is ΣnCr = 2n. 4. 5 heads: 252/1024 = 0.246. 5. 0.246, 0.176, 0.112, 0.0796: it decreases, roughly like 1/√n.

20 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 10 minutes)

(a) Expand (2 + x)4. [3]

(b) Hence find the exact value of 2.14. [2]

Markscheme

Part	Answer	Marks
(a)	16 + 32x + 24x2 + 8x3 + x4	A3
(b)	x = 0.1: 16 + 3.2 + 0.24 + 0.008 + 0.0001	M1
	19.4481	A1

Examiner's comments. Substituting x = 0.1 in the full expansion gives the exact value.

Question 2 (Paper 1 style) (about 10 minutes)

Consider the expansion of (3x2 − 1/x)9.

(a) Write down the number of terms. [1]

(b) Find the term independent of x. [4]

Markscheme

Part	Answer	Marks
(a)	10	A1
(b)	9Cr (3x2)9−r(−1/x)r	M1
	18 − 3r = 0, r = 6	A1
	84 × 27 × 1	M1
	2268	A1

Examiner's comments. (−1)6 = 1, so the term is positive.

Question 3 (Paper 1 style) (about 8 minutes)

The coefficient of x2 in the expansion of (1 + kx)8 is 252. Find the possible values of k. [4]

Markscheme

Part	Answer	Marks
	8C2 = 28	A1
	28k2 = 252	M1
	k2 = 9	A1
	k = ±3	A1

Examiner's comments. Both signs are valid.

21 Common misconceptions and links

Students often think…	Actually…
(2x)3 = 2x3	(2x)3 = 8x3
the rth term uses nCr	the (r + 1)th term uses nCr
(a − b)n has all positive terms	signs alternate
0! = 0	0! = 1

How 1.9 connects to the rest of the course

Topic	Connection
4.7 Binomial distribution	probabilities
1.3 Geometric sequences	compound interest
4.5 Probability	counting outcomes
5.1 Limits	approximations

22 Check your understanding

Section A: multiple choice

1. 6C2 =

A. 15 B. 12 C. 30 D. 36

2. (a + b)8 has

A. 8 terms B. 16 terms C. 9 terms D. 7 terms

3. Row sums of Pascal's triangle are

A. triangular numbers B. powers of 2 C. factorials D. primes

4. The constant term of (x + 2/x)2 is

A. 2 B. 0 C. 8 D. 4

5. nCr = nC(n − r) is

A. Pascal's rule B. symmetry C. the row sum D. false

Section B: structured

6. Find the coefficient of x3 in (2x − 5)6. [3]

7. Expand (1 + x)6 and use it to estimate 1.016 to 6 d.p. [3]

Answers

1. A 2. C 3. B 4. D 5. B

6. 6C3 × 23 × (−5)3 = 20 × 8 × (−125) = −20 000.

7. 1.061 520

23 Summary and topic map

Idea	Remember
Pascal's triangle	each entry is the sum of the two above
Factorial	n! with 0! = 1
nCr	n!/(r!(n − r)!): choices
Theorem	(a + b)n = ΣnCr an−rbr
General term	term r + 1
Applications	counting, probability, approximations

Before the exam, make sure you…

•raise the whole term to the power
•keep negative signs
•solve for r from the power of x
•use nCr on the GDC

The Binomial Theorem on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for The Binomial Theorem.

24 Answers to practice exercises

Check your working against these answers. Full methods are expected in the exam.

Practice exercise 1

1. x2 − 6x + 9

2. 8x3 + 12x2 + 6x + 1

3. 13

Practice exercise 2

1. 1 9 36 84 126 126 84 36 9 1

2. y5 + 10y4 + 40y3 + 80y2 + 80y + 32

3. 210 = 1024

Practice exercise 3

1. 5040

2. 190

3. 720

Practice exercise 4

1. 210

2. 220

3. n = 9

Practice exercise 5

1. both 35

2. 15 + 20 = 35 = 7C3

3. 128

Practice exercise 6

1. 128 + 448x + 672x2

2. 12Cr (3x)12−r 5r

Practice exercise 7

1. 1 − 5x + 10x2 − 10x3 + 5x4 − x5

2. x3/8 + 3x2 + 24x + 64

3. x4 − 4x2 + 6 − 4/x2 + 1/x4

Practice exercise 8

1. 15 × 81 = 1215

2. 21 × (−32) = −672

3. 10C4 26x4 = 13 440x4

Practice exercise 9

1. 20

2. 8C4 × 16 × 1 = 1120

3. 5 − 2r = 0 has no integer solution

Practice exercise 10

1. 10 × 4 × p3 = 720: p3 = 18, p = 2.62

2. 6C4 k4 = 240: k4 = 16, k = 2

Practice exercise 11

1. (1 + 2x + x2)(1 − 4x + 6x2 − 4x3 + ...): −4 + 12 − 4 = 4

2. 2(−18) + 1 = −35

Practice exercise 12

1. 1 − 0.08 + 0.0028 = 0.9228

2. 32 + 80(0.01) + 80(0.0001) = 32.808

Practice exercise 13

1. 2 598 960

2. 12C6 = 924

Practice exercise 14

1. 10 × (1/36)(125/216) = 0.161

2. they are the terms of (q + p)n = 1n = 1

Practice exercise 15

1. 64

2. triangular numbers

Practice exercise 16

1. 1 + 0.12 + 0.0054 = 1.1254

2. the neglected higher-power terms are no longer small

Practice exercise 17

1. 5 200 300

2. 15C7 = 15C8 = 6435

Practice exercise 18

1. 56 − 56 = 0

2. 6C3 × 27x3 × (−1)3 = −540x3

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