DP Maths AA SL - Number and Algebra - Binomial Theorem

DP Maths AA SL - Number and Algebra - Binomial Theorem - Full notes

RevisionPrep

DP Maths AA SL | Binomial Theorem

Binomial Theorem

This topic contains 1 note:

1.6 Proof and Binomial Theorem

DP MATHEMATICS AA · TOPIC 1: NUMBER AND ALGEBRA

1.6

Proof and the Binomial Theorem

Full notes for SL · Syllabus SL 1.6, 1.9

Contents

1	What is a proof?
2	Notation and the language of proof
3	Writing deductive proofs
4	Proving identities
5	Pascal's triangle and binomial expansions
6	Factorials and binomial coefficients
7	The binomial theorem
8	Finding a specific term or coefficient
9	Products of expansions and approximations
10	Direct proofs with integers
11	Proving algebraic identities
12	Proofs involving inequalities
13	Errors in proof
14	Patterns in Pascal's triangle
15	Expansions with fractions and negative terms
16	Finding coefficients with x in both terms
17	Products of two expansions
18	Approximations using the binomial theorem
19	Binomial coefficients in counting and probability
20	Mixed review and exam practice
21	Exam-style questions with markscheme
22	Common misconceptions and links
23	Check your understanding
24	Summary and topic map
25	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 Quick check.

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
explain the difference between checking examples and proving a statement
use the notation ≡, ⇒, ⇔, ∀, ∈, ℤ, ℚ, ℝ correctly
write deductive proofs about integers (odd, even, consecutive, multiples)
prove algebraic identities working from one side to the other
expand (a + b)n using Pascal's triangle and the binomial theorem
evaluate nCr with and without technology
find a specific term, coefficient or constant term
use expansions to find coefficients in products and to approximate

This booklet covers SL 1.6, 1.9, which is common to SL and HL.

Guiding questions

What makes a mathematical argument a proof? How can large powers of a binomial be expanded efficiently, and why do Pascal's triangle and nCr give the same numbers?

How this topic is examined

Proof is assessed through 'show that' and 'prove that' questions on both papers; clear communication earns marks. The binomial theorem appears very frequently on Paper 1, usually asking for a specific term, a coefficient, or an unknown constant given a coefficient.

In a proof, start from one side and reach the other, writing each step on a new line, define every variable with its set (n ∈ ℤ), and finish with a concluding statement.

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
Solve	find the values of the unknowns, using algebra or technology as allowed
Show that	obtain the given result; every step must be shown and the result must not be assumed
Hence / Hence or otherwise	use the previous result (or another valid method)
Sketch	a diagram showing the general shape and key features, labelled
Prove	a formal, rigorous argument from accepted facts
Justify / Explain	give valid reasons or evidence

How to use these notes

Each section teaches one part of the syllabus. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Why it works	a derivation or proof of a key result, so you understand where formulas come from
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 on Paper 2
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); examples are graded basic, standard and exam level
Practice exercise	graded questions for each section; full answers are at the end of the booklet
Quick check	short questions to check the key ideas

1 What is a proof?

Showing that something is always true.

Think about it

The numbers 3, 5, 7, 11 and 13 are all prime. Does checking many odd numbers prove that every odd number greater than 1 is prime?

In everyday life we often accept a claim after seeing many examples. In mathematics this is not enough: a proof is a logical argument that shows a statement is true in every possible case, starting from definitions and results already known to be true.

Checking examples can suggest a pattern (this is called a conjecture), but cannot prove it, because there are infinitely many cases. However, a single counterexample is enough to show a general statement is false: 9 is odd and not prime, so the statement in the box above is false.

Famous conjectures show why proof matters. The expression n2 + n + 41 gives primes for n = 0, 1, 2, ..., 39, but fails at n = 40 (402 + 40 + 41 = 1681 = 412). A pattern that holds for forty cases still was not true in general.

Key point

Examples can suggest a result; only a proof establishes it. One counterexample is enough to disprove a general claim.

2 Notation and the language of proof

Symbols you must read and write correctly.

Symbol	Meaning	Example
⇒	implies	x = 2 ⇒ x2 = 4
⇔	if and only if (implies in both directions)	x + 3 = 5 ⇔ x = 2
≡	is identically equal to (true for all values)	(x + 1)2 ≡ x2 + 2x + 1
∀	for all	∀x ∈ ℝ, x2 ≥ 0
∈	is an element of	3 ∈ ℤ
ℤ, ℕ, ℚ, ℝ	integers, natural numbers, rationals, reals	√2 ∈ ℝ but √2 ∉ ℚ

Note that x2 = 4 does not imply x = 2 (x could be −2), so the arrow goes only one way in that example. Using ⇔ when only ⇒ is valid is a common logical error.

Useful general forms (with n, m ∈ ℤ): an even number 2n; an odd number 2n + 1; consecutive integers n, n + 1, n + 2; consecutive even numbers 2n, 2n + 2; a multiple of k kn; a number that leaves remainder r when divided by k: kn + r. To refer to two unrelated odd numbers, use two different letters: 2m + 1 and 2n + 1.

Practice exercise 1

1. Write a general expression for three consecutive odd numbers.

2. Is it true that x2 = 9 ⇔ x = 3? Explain.

3. Give a counterexample to 'all prime numbers are odd'.

4. Write a general number that leaves remainder 2 when divided by 5.

3 Writing deductive proofs

Clear structure, one step per line.

A deductive proof moves step by step from what is known to what is to be shown. A good proof: (1) defines the variables and their sets; (2) writes the given information algebraically; (3) manipulates it with valid algebra, one step per line; (4) interprets the result; (5) ends with a clear concluding statement linked back to the question.

WORKED EXAMPLE 1 Sum of consecutive integers (basic)

Prove that the sum of any three consecutive integers is a multiple of 3.

Solution

Let the integers be n, n + 1, n + 2, where n ∈ ℤM1

Sum = 3n + 3 = 3(n + 1)A1

Since n + 1 ∈ ℤ, the sum is a multiple of 3R1

WORKED EXAMPLE 2 Squares of odd numbers (standard)

Prove that the square of any odd number is odd.

Solution

Let the odd number be 2n + 1, n ∈ ℤM1

(2n + 1)2 = 4n2 + 4n + 1 = 2(2n2 + 2n) + 1A1

2n2 + 2n ∈ ℤ, so the square has the form 2k + 1 and is oddR1

WORKED EXAMPLE 3 Difference of squares of consecutive odd numbers (exam level)

Prove that the difference between the squares of two consecutive odd numbers is a multiple of 8.

Solution

Let the numbers be 2n + 1 and 2n + 3, n ∈ ℤM1

(2n + 3)2 − (2n + 1)2 = (4n2 + 12n + 9) − (4n2 + 4n + 1)M1

= 8n + 8 = 8(n + 1)A1

n + 1 ∈ ℤ, so the difference is a multiple of 8R1

WORKED EXAMPLE 4 Using a counterexample (standard)

Disprove: 'For all real numbers x, x3 ≥ x.'

Solution

Take x = ½: x3 = 1/8 < ½A1

So the statement is falseR1

Common mistake

Starting from the result to be proved and working backwards to something true. This assumes what you are trying to prove; begin from the given facts instead.

Practice exercise 2

1. Prove that the sum of two odd numbers is even.

2. Prove that the product of two consecutive integers is even.

3. Prove that the sum of four consecutive integers is never a multiple of 4.

4. Disprove: 'if n is prime, then 2n + 1 is prime'.

5. Prove that (n + 1)2 − (n − 1)2 is a multiple of 4 for all n ∈ ℤ.

4 Proving identities

Work on one side until it becomes the other.

An identity is true for all values of the variable, indicated by ≡. To prove an identity, start with one side (usually the more complicated one, often the left-hand side, LHS), manipulate it, and show that it equals the other side (RHS). Do not move terms across the ≡ sign as if solving an equation, and do not work on both sides at once.

An equation, by contrast, is true only for particular values: x2 = 4 is an equation; (x + 2)2 ≡ x2 + 4x + 4 is an identity.

WORKED EXAMPLE 5 An algebraic identity (basic)

Prove that (n + 3)2 − (n − 3)2 ≡ 12n.

Solution

LHS = (n2 + 6n + 9) − (n2 − 6n + 9)M1

= 12n = RHSA1

WORKED EXAMPLE 6 A fraction identity (standard)

Show that 1/(x − 1) − 1/(x + 1) ≡ 2/(x2 − 1).

Solution

LHS = ((x + 1) − (x − 1)) ÷ ((x − 1)(x + 1))M1

= 2 ÷ (x2 − 1) = RHSA1

WORKED EXAMPLE 7 Finding constants in an identity (exam level)

Find a, b and c such that 2x2 + 12x + 7 ≡ a(x + b)2 + c.

Solution

Expand: a(x + b)2 + c = ax2 + 2abx + ab2 + cM1

Compare coefficients: a = 2; 2ab = 12 so b = 3; ab2 + c = 7 so c = −11A1 A1

a = 2, b = 3, c = −11A1

Practice exercise 3

1. Prove that (a + b)2 − (a − b)2 ≡ 4ab.

2. Show that x/(x + 2) + 2/(x − 2) ≡ (x2 + 4)/(x2 − 4).

3. Find p and q such that x2 − 8x + 20 ≡ (x − p)2 + q, and hence show x2 − 8x + 20 > 0 for all x.

5 Pascal's triangle and binomial expansions

Patterns in the coefficients.

A binomial is an expression with two terms, such as a + b or 2x − 3. Expanding powers of (a + b) by repeated multiplication reveals a pattern: (a + b)2 = a2 + 2ab + b2; (a + b)3 = a3 + 3a2b + 3ab2 + b3; (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4.

The coefficients form Pascal's triangle, in which every row starts and ends with 1 and each inner number is the sum of the two numbers above it. This happens because each term of (a + b)n+1 = (a + b)(a + b)n comes from two terms of the row above.

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

n	Coefficients in (a + b)n	Sum
0	1	1
1	1 1	2
2	1 2 1	4
3	1 3 3 1	8
4	1 4 6 4 1	16
5	1 5 10 10 5 1	32
6	1 6 15 20 15 6 1	64

Notice that each row sums to 2n: substituting a = b = 1 into (a + b)n gives 2n.

WORKED EXAMPLE 8 Using Pascal's triangle (basic)

Expand (x + 3)4.

Solution

Coefficients 1, 4, 6, 4, 1M1

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

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

WORKED EXAMPLE 9 A negative term (standard)

Expand (2x − 1)4.

Solution

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

(2x)4 + 4(2x)3(−1) + 6(2x)2(−1)2 + 4(2x)(−1)3 + (−1)4M1

= 16x4 − 32x3 + 24x2 − 8x + 1A1

With a negative second term, the signs alternate.

Practice exercise 4

1. Write down row 7 of Pascal's triangle.

2. Expand (1 + x)5.

3. Expand (3 − x)3.

4. Expand (x + 2/x)4 and simplify each term.

6 Factorials and binomial coefficients

Counting choices.

The factorial of a non-negative integer is n! = n × (n − 1) × ... × 2 × 1, with 0! = 1 by definition. Factorials grow very fast: 10! = 3 628 800.

The binomial coefficient nCr (read 'n choose r') counts the number of ways of choosing r objects from n different objects, ignoring order. It equals the entry in position r (counting from r = 0) of row n of Pascal's triangle:

Why do choices give binomial coefficients? Expanding (a + b)n means choosing either a or b from each of the n brackets. The term an−rbr arises each time b is chosen from exactly r of the brackets, and there are nCr ways to make that choice.

Properties: nC0 = nCn = 1; nC1 = n; nCr = nC(n − r) (symmetry); and nCr + nC(r + 1) = (n + 1)C(r + 1) (Pascal's rule).

WORKED EXAMPLE 10 Evaluating without a calculator (basic)

Find 8C3.

Solution

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

= 56A1

WORKED EXAMPLE 11 Solving with binomial coefficients (exam level)

Solve nC2 = 45.

Solution

n(n − 1)/2 = 45, so n2 − n − 90 = 0M1

(n − 10)(n + 9) = 0A1

n ∈ ℕ, so n = 10A1

GDC tip

Use the nCr function (in the probability menu) for binomial coefficients. On Paper 1, cancel factorials by hand: nCr = n(n − 1)...(n − r + 1)/r!.

Practice exercise 5

1. Evaluate 6C2 and 10C7.

2. Simplify (n + 1)!/(n − 1)!.

3. Solve nC2 = 28.

4. Show that 5C2 + 5C3 = 6C3.

7 The binomial theorem

A formula for every expansion.

For n ∈ ℕ, the binomial theorem gives the full expansion:

The general term, called the (r + 1)th term, is:

WORKED EXAMPLE 12 Using the theorem (standard)

Find the first three terms, in ascending powers of x, of (2 + 3x)6.

Solution

26 + 6C1 × 25(3x) + 6C2 × 24(3x)2M1

= 64 + 6 × 32 × 3x + 15 × 16 × 9x2A1

= 64 + 576x + 2160x2A1

Common mistake

Forgetting to raise the coefficient of x to the power: (3x)2 = 9x2, not 3x2.

Practice exercise 6

1. Find the first three terms of (1 − 2x)8.

2. Find the first three terms of (3 + x/2)5 in ascending powers of x.

3. Write down the term in x4 in (1 + x)9.

8 Finding a specific term or coefficient

Write the general term, then solve for r.

To find the term in a particular power of x: write the general term; collect all powers of x into a single power xf(r); set f(r) equal to the required power and solve for r; then evaluate. The constant term (term independent of x) is found by setting the power of x equal to 0.

WORKED EXAMPLE 13 A specific coefficient (standard)

Find the coefficient of x3 in the expansion of (3 − 2x)7.

Solution

General term: 7Cr 37−r(−2x)r; need r = 3M1

7C3 × 34 × (−2)3 = 35 × 81 × (−8)A1

= −22 680A1

WORKED EXAMPLE 14 Term independent of x (exam level)

Find the constant term in the expansion of (x2 + 2/x)6.

Solution

General term: 6Cr (x2)6−r(2/x)r = 6Cr 2r x12−3rM1

12 − 3r = 0, so r = 4A1

6C4 × 24 = 15 × 16 = 240A1

WORKED EXAMPLE 15 Finding an unknown constant (exam level)

The coefficient of x2 in (1 + kx)5 is 40. Find the possible values of k.

Solution

5C2 k2 = 40M1

10k2 = 40, k2 = 4A1

k = ±2A1

Practice exercise 7

1. Find the coefficient of x5 in (x − 2)8.

2. Find the constant term in (2x − 1/x)6.

3. Find the term in x2 in (x + 3/x)6.

4. The coefficient of x3 in (2 + ax)5 is 1080. Find a.

5. Find n if the coefficient of x2 in (1 + 3x)n is 252.

9 Products of expansions and approximations

Combining expansions; using the first few terms.

To find a coefficient in a product such as (1 + 2x)(1 − x)6, list the terms of each factor that can multiply to give the required power, then add their products.

When x is small, higher powers of x are tiny, so the first few terms of a binomial expansion give a good approximation: 1.026 = (1 + 0.02)6 ≈ 1 + 6(0.02) + 15(0.02)2 = 1.126 (actual 1.126 16...).

WORKED EXAMPLE 16 Coefficient in a product (exam level)

Find the coefficient of x2 in (1 + 2x)(1 − x)6.

Solution

(1 − x)6 = 1 − 6x + 15x2 − ...A1

x2 terms: 1 × 15x2 + 2x × (−6x)M1

= 15 − 12 = 3A1

WORKED EXAMPLE 17 Approximation (standard)

Use the first three terms of (1 − x/2)8 to estimate 0.998.

Solution

(1 − x/2)8 ≈ 1 − 4x + 7x2M1

0.99 = 1 − x/2 gives x = 0.02M1

≈ 1 − 0.08 + 0.0028 = 0.9228A1

Practice exercise 8

1. Find the coefficient of x3 in (2 − x)(1 + x)5.

2. Use (1 + x)10 to estimate 1.00310 to 6 d.p.

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

10 Direct proofs with integers

Even, odd and consecutive numbers.

To prove statements about integers, represent them algebraically: an even number is 2n, an odd number 2n + 1, consecutive integers n and n + 1, and a multiple of 3 is 3n (n ∈ ℤ). Then manipulate the expression to show the required form.

WORKED EXAMPLE 18 Sum of consecutive odd numbers (standard)

Prove that the sum of two consecutive odd numbers is a multiple of 4.

Solution

(2n + 1) + (2n + 3) = 4n + 4M1

= 4(n + 1), a multiple of 4A1 AG

WORKED EXAMPLE 19 Difference of squares (exam level)

Prove that the difference between the squares of two consecutive integers is always odd.

Solution

(n + 1)2 − n2 = 2n + 1M1 A1

2n + 1 is odd for all integers nR1 AG

Practice exercise 9

1. Prove that the product of two odd numbers is odd.

2. Prove that the sum of three consecutive integers is divisible by 3.

3. Prove that n2 + n is always even.

11 Proving algebraic identities

Left-hand side to right-hand side.

An identity (≡) holds for all values of the variable. Prove it by starting with one side (usually the more complicated) and transforming it into the other, step by step. Do not start by assuming the result, and do not work on both sides at once unless each step is reversible.

WORKED EXAMPLE 20 Expanding (standard)

Prove that (a + b)2 − (a − b)2 ≡ 4ab.

Solution

LHS = a2 + 2ab + b2 − a2 + 2ab − b2M1

= 4ab = RHSA1 AG

WORKED EXAMPLE 21 Fractions (exam level)

Prove that 1/n − 1/(n + 1) ≡ 1/(n(n + 1)).

Solution

LHS = ((n + 1) − n)/(n(n + 1))M1

= 1/(n(n + 1)) = RHSA1 AG

Practice exercise 10

1. Prove that (x + 3)2 − (x − 3)2 ≡ 12x.

2. Prove that n3 − n ≡ (n − 1)n(n + 1), and hence that n3 − n is always divisible by 6.

12 Proofs involving inequalities

Using squares.

Many inequalities follow from the fact that a square is never negative. For example, completing the square shows that x2 − 6x + 10 = (x − 3)2 + 1 > 0 for all x.

WORKED EXAMPLE 22 Completing the square (standard)

Prove that x2 + 4x + 7 > 0 for all real x.

Solution

x2 + 4x + 7 = (x + 2)2 + 3M1 A1

(x + 2)2 ≥ 0, so the expression ≥ 3 > 0R1 AG

WORKED EXAMPLE 23 Arithmetic-geometric mean (exam level)

Prove that for positive a and b, (a + b)/2 ≥ √(ab).

Solution

(√a − √b)2 ≥ 0M1

a − 2√(ab) + b ≥ 0A1

so (a + b)/2 ≥ √(ab)AG

Practice exercise 11

1. Prove that a2 + b2 ≥ 2ab for all real a, b.

2. Prove that 2x2 − 4x + 5 > 0 for all x.

13 Errors in proof

What does not count as a proof.

Checking a few examples does not prove a general statement (but one counterexample disproves it). Starting from the result and deriving a true statement is not valid unless every step is reversible. Dividing by an expression that could be zero is a common hidden error.

WORKED EXAMPLE 24 Spot the error (standard)

A student 'proves' 1 = 2: let a = b; then a2 = ab, so a2 − b2 = ab − b2, (a + b)(a − b) = b(a − b), a + b = b, 2b = b, 2 = 1. Find the error.

Solution

Dividing by a − b, which is 0R1

The step from (a + b)(a − b) = b(a − b) to a + b = b is invalidR1

Practice exercise 12

1. Give a counterexample to 'n2 + n + 41 is prime for every positive integer n'.

2. Why does checking n = 1, 2, ..., 100 not prove a statement for all n?

14 Patterns in Pascal's triangle

Symmetry, sums and diagonals.

Each row of Pascal's triangle is symmetric (nCr = nC(n − r)), each entry is the sum of the two above (nCr + nC(r + 1) = (n + 1)C(r + 1)), and the entries in row n sum to 2n. The diagonals contain the natural numbers, the triangular numbers and the tetrahedral numbers.

WORKED EXAMPLE 25 Row sum (standard)

Show that nC0 + nC1 + ... + nCn = 2n.

Solution

Substitute a = b = 1 in (a + b)n = Σ nCr an−rbrM1

2n = Σ nCrAG

WORKED EXAMPLE 26 Alternating sum (exam level)

Find nC0 − nC1 + nC2 − ... + (−1)nnCn for n ≥ 1.

Solution

Substitute a = 1, b = −1M1

(1 − 1)n = 0A1

Practice exercise 13

1. Evaluate 6C0 + 6C1 + ... + 6C6.

2. Verify Pascal's rule for 7C3 + 7C4.

3. Find the sum of the even-position entries of row 8 (8C0 + 8C2 + ... + 8C8).

15 Expansions with fractions and negative terms

Careful with signs and powers.

When terms such as 2/x or −x/3 appear, raise the whole term (including its sign and coefficient) to the power. Write the general term and simplify the powers of x before substituting.

WORKED EXAMPLE 27 Fractional term (standard)

Find the first four terms of (2 − x/4)6.

Solution

64 + 6(32)(−x/4) + 15(16)(x2/16) + 20(8)(−x3/64)M1

= 64 − 48x + 15x2 − 2.5x3A1 A1

Practice exercise 14

1. Expand (1 − 2x)5.

2. Find the first three terms of (3 + x/3)4.

16 Finding coefficients with x in both terms

Solve for r.

For (axp + b/xq)n, the general term has power of x equal to p(n − r) − qr. Set this equal to the required power and solve for r; if r is not a whole number between 0 and n, no such term exists.

WORKED EXAMPLE 28 A coefficient with x in both terms (exam level)

Find the coefficient of x6 in (x2 + 3/x)9.

Solution

Power: 2(9 − r) − r = 18 − 3r = 6, r = 4M1 A1

9C4 × 34 = 126 × 81 = 10 206A1

WORKED EXAMPLE 29 No such term (standard)

Show that (x + 1/x2)7 has no constant term.

Solution

Power 7 − 3r = 0 gives r = 7/3M1

Not an integer: no constant termR1

Practice exercise 15

1. Find the constant term in (2x − 1/x2)6.

2. Find the coefficient of x5 in (x2 − 2/x)7.

17 Products of two expansions

Combine contributions.

To find a coefficient in a product such as (2 + x)(1 − x)8, expand each factor only as far as needed and add the contributions that give the required power.

WORKED EXAMPLE 30 Product (exam level)

Find the coefficient of x3 in (2 + x)(1 − x)8.

Solution

(1 − x)8 = 1 − 8x + 28x2 − 56x3 + ...A1

2(−56) + 1(28)M1

= −84A1

Practice exercise 16

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

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

18 Approximations using the binomial theorem

Small x.

When x is small, the first few terms of (1 + x)n give a good approximation. Choose x so that the bracket matches the number to be estimated.

WORKED EXAMPLE 31 Estimate (standard)

Use the expansion of (1 + x)8 to estimate 1.0038 to 6 decimal places.

Solution

1 + 8(0.003) + 28(0.003)2 + 56(0.003)3M1

= 1 + 0.024 + 0.000252 + 0.0000015 = 1.024254A1

Practice exercise 17

1. Estimate 0.985 using three terms.

2. Estimate 2.014 using (2 + x)4.

19 Binomial coefficients in counting and probability

Same numbers, new contexts.

nCr counts the ways to choose r items from n. In probability, the chance of exactly r successes in n independent trials with probability p is nCr pr(1 − p)n−r, which are the terms of (q + p)n.

WORKED EXAMPLE 32 Coins (standard)

A fair coin is tossed 8 times. Find P(exactly 5 heads).

Solution

8C5 (0.5)8M1

= 56/256 = 0.219A1

Practice exercise 18

1. How many ways can a committee of 4 be chosen from 10 people?

2. A die is rolled 4 times. Find P(exactly one six).

20 Mixed review and exam practice

Bringing the ideas together.

WORKED EXAMPLE 33 Paper 1 question (exam level)

The coefficient of x2 in (1 + ax)n is 45 and the coefficient of x is 10. Find a and n.

Solution

na = 10; n(n − 1)a2/2 = 45M1

(n − 1)a = 9, so a = 1A1

a = 1, n = 10A1

WORKED EXAMPLE 34 Proof question (exam level)

Prove that the square of any odd number is one more than a multiple of 8.

Solution

(2n + 1)2 = 4n2 + 4n + 1 = 4n(n + 1) + 1M1 A1

n(n + 1) is even, so 4n(n + 1) is a multiple of 8R1 AG

Practice exercise 19

1. Find the term in x4 in (3x − 2)6.

2. Prove that the sum of any five consecutive integers is divisible by 5.

21 Exam-style questions with markscheme

Attempt each question under timed conditions before reading the markscheme.

Question 1 (Paper 1 style) (about 15 minutes)

(a) Prove that the difference between the squares of any two consecutive integers is odd. [3]

(b) Find the coefficient of x4 in the expansion of (2x − 3)6. [3]

Markscheme

Part	Answer	Marks
(a)	(n + 1)2 − n2	M1
	= 2n + 1	A1
	2n + 1 is odd for n ∈ ℤ	R1
(b)	6C2 (2x)4(−3)2	M1
	15 × 16 × 9	A1
	2160	A1
(c)	5C2 k2 = 40	M1
	k2 = 4	A1
	k = ±2	A1

Examiner's comments. In (b), the most common error was choosing r = 4 instead of r = 2: the power of x comes from (2x)6−r.

Question 2 (Paper 1 style) (about 12 minutes)

Consider the expansion of (x + a/x2)9, a > 0.

(a) Write down the general term. [2]

(b) The constant term is 5376. Find a. [4]

Markscheme

Part	Answer	Marks
(a)	9Cr x9−r(a/x2)r	A1
	= 9Cr ar x9−3r	A1
(b)	9 − 3r = 0, r = 3	M1 A1
	84a3 = 5376	M1
	a = 4	A1

Examiner's comments. Collecting powers of x into a single exponent is the key step.

22 Common misconceptions and links

Students often think…	Actually…
checking examples proves a statement	a proof must cover every case
the rth term uses nCr	the (r + 1)th term uses nCr
(a − b)n has all positive terms	signs alternate
⇒ and ⇔ mean the same	⇔ requires both directions

How 1.6 connects to the rest of the course

Topic	Connection
1.10 Counting principles (HL)	nCr, permutations, extended binomial
1.15 Further proof (HL)	induction and contradiction
4.7 Binomial distribution	nCr in probability
2.12 Polynomials (HL)	expansions

23 Check your understanding

Section A: multiple choice

1. 5C2 =

A. 10 B. 20 C. 5 D. 25

2. The number of terms in (x + y)7 is

A. 7 B. 6 C. 8 D. 14

3. A counterexample shows a statement is

A. true B. false C. an identity D. proven for all cases

4. The constant term of (x + 1/x)4 is

A. 4 B. 1 C. 0 D. 6

5. 0! =

A. 0 B. 1 C. undefined D. ∞

6. ≡ means

A. identically equal for all values B. approximately equal C. implies D. not equal

Section B: structured

7. Prove that the product of two odd numbers is odd. [3]

8. Expand (1 + 2x)5 fully. [3]

9. Find the term in x5 in (x − 2)8. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A

7. (2m + 1)(2n + 1) = 2(2mn + m + n) + 1: odd.

8. 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5.

9. 8C3 x5(−2)3 = −448x5.

24 Summary and topic map

Idea	Remember
Proof	general argument; counterexample disproves
Notation	⇒, ⇔, ≡, ∀, ∈, ℤ, ℝ
Identities	LHS to RHS; compare coefficients
Pascal's triangle	each entry is the sum of the two above
nCr	n!/(r!(n − r)!)
General term	nCr an−rbr is term r + 1
Products	add contributions from each pair of terms

Before the exam, make sure you…

•define variables with their sets
•end proofs with a conclusion
•include signs and coefficients in powers
•solve for r from the power of x

Proof and 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 1. Topic map for Proof and the Binomial Theorem.

25 Answers to practice exercises

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

Practice exercise 1

1. 2n + 1, 2n + 3, 2n + 5

2. No: x2 = 9 ⇒ x = ±3, so only x = 3 ⇒ x2 = 9 holds

3. 2

4. 5n + 2, n ∈ ℤ

Practice exercise 2

1. (2m + 1) + (2n + 1) = 2(m + n + 1)

2. n(n + 1): one of the two is even, so the product is even

3. 4n + 6 = 4(n + 1) + 2: remainder 2 when divided by 4

4. n = 7: 2n + 1 = 15, not prime

5. = 4n

Practice exercise 3

1. expand and simplify the LHS

2. common denominator (x + 2)(x − 2): numerator x(x − 2) + 2(x + 2) = x2 + 4

3. p = 4, q = 4; (x − 4)2 ≥ 0 so the expression ≥ 4 > 0

Practice exercise 4

1. 1 7 21 35 35 21 7 1

2. 1 + 5x + 10x2 + 10x3 + 5x4 + x5

3. 27 − 27x + 9x2 − x3

4. x4 + 8x2 + 24 + 32/x2 + 16/x4

Practice exercise 5

1. 15; 120

2. n(n + 1)

3. n = 8

4. 10 + 10 = 20 = 6C3

Practice exercise 6

1. 1 − 16x + 112x2

2. 243 + 202.5x + 67.5x2

3. 126x4

Practice exercise 7

1. −448

2. −160

3. 135x2

4. a = 3

5. 9n(n − 1)/2 = 252: n = 8

Practice exercise 8

1. 2 × 10 − 10 = 10

2. 1 + 0.03 + 0.000 405 = 1.030 405 (actual 1.030 408)

3. (x3 + 3x + 3/x + 1/x3)(x + 2): only 3x × 2 gives an x term, so the coefficient is 6

Practice exercise 9

1. (2a + 1)(2b + 1) = 2(2ab + a + b) + 1

2. n + (n + 1) + (n + 2) = 3(n + 1)

3. n(n + 1) is a product of consecutive integers, one of which is even

Practice exercise 10

1. expand: 12x

2. factorize; product of three consecutive integers is divisible by 2 and 3

Practice exercise 11

1. (a − b)2 ≥ 0

2. 2(x − 1)2 + 3 > 0

Practice exercise 12

1. n = 40: 1681 = 412

2. the statement could fail for a larger n

Practice exercise 13

1. 64

2. 35 + 35 = 70 = 8C4

3. 128

Practice exercise 14

1. 1 − 10x + 40x2 − 80x3 + 80x4 − 32x5

2. 81 + 36x + 6x2

Practice exercise 15

1. 6 − 3r = 0: r = 2: 15 × 16 × 1 = 240

2. 14 − 3r = 5: r = 3: 35 × (−8) = −280

Practice exercise 16

1. (2 − x)5: 32 − 80x + 80x2: 80 + 3(−80) = −160

2. 1 × 1 (from x × 1) + (1/x)(6x2): 1 + 6 = 7

Practice exercise 17

1. 1 − 0.1 + 0.004 = 0.904

2. 16 + 0.32 + 0.0024 = 16.3224

Practice exercise 18

1. 210

2. 4 × (1/6)(5/6)3 = 0.386

Practice exercise 19

1. 15 × 81 × 4 x4 = 4860x4

2. 5n + 10 = 5(n + 2)

revisionprep.comIB-style notes · question bank · mock papers