← IB Mathematics: Analysis & Approaches · HL · IB Maths AA revision

IB Diploma Mathematics: Analysis & Approaches HL · Number & Algebra

Mini-Lesson

Number & Algebra

8 missions. Guess, learn, then check. Your place is saved.

Mission 1How many cells laid end to end would cross the Milky Way?
Mission 220 seats in row 1, two more in each row: how many seats in 30 rows?
Mission 3Double the rice on each chessboard square. How much on all 64?
Mission 4£10,000 at 4% for 10 years: what does monthly compounding add?
Mission 5A magnitude 8 earthquake versus a magnitude 6: how much bigger?
Mission 6Why do the digits of $1 1^{4} = 14641$ 11⁴ = 14641 form a row of Pascal's triangle?
Mission 7How many ways can 8 runners finish 1st, 2nd and 3rd?
Mission 8Three equations, three unknowns: how can there be no solution?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

How many cells laid end to end would cross the Milky Way?

Make a guess: How many cells laid end to end would cross the Milky Way?

Number and algebra · standard form and approximation

Standard form

A number is in standard form when it is written $a \times 1 0^{k}$ a × 10^k with $1 \leq a < 10$ 1 ≤ a < 10 and $k$ k an integer.
Large numbers have $k > 0$ k > 0: $305000 = 3.05 \times 1 0^{5}$ 305 000 = 3.05 × 10⁵.
Small numbers have $k < 0$ k < 0: $0.000 72 = 7.2 \times 1 0^{- 4}$ 0.000 72 = 7.2 × 10⁻⁴.
Calculator notation such as 5.2E30 earns no credit in IB exams. Write $5.2 \times 1 0^{30}$ 5.2 × 10³⁰.

Quick check

?Which is

0.000 304

written in standard form?

Number and algebra · standard form and approximation

Calculating in standard form

Multiply: multiply the $a$ a parts and add the indices. $(3 \times 1 0^{4}) \times (5 \times 1 0^{6}) = 15 \times 1 0^{10}$ (3 × 10⁴) × (5 × 10⁶) = 15 × 10¹⁰.
Divide: divide the $a$ a parts and subtract the indices: $\frac{1 0 ^{m}}{1 0 ^{n}} = 1 0^{m - n}$ (10^m)/10ⁿ = 10^(m−n).
Then adjust so $1 \leq a < 10$ 1 ≤ a < 10: $15 \times 1 0^{10} = 1.5 \times 1 0^{11}$ 15 × 10¹⁰ = 1.5 × 10¹¹.
Add or subtract: first rewrite both numbers with the same power of 10.
Give final answers exact or to 3 significant figures, the IB default.

Worked example Number and algebra · standard form and approximation

Dividing in standard form

Question

A hard drive holds

2.4 \times 1 0^{12}

bytes. One photo uses

3.2 \times 1 0^{6}

bytes. How many photos fit on the drive? Give your answer in standard form.

1
Write the division: $\frac{2.4 \times 1 0 ^{12}}{3.2 \times 1 0 ^{6}}$ (2.4 × 10¹²)/(3.2 × 10⁶)
2
Divide the numbers: $2.4 \div 3.2 = 0.75$ 2.4 ÷ 3.2 = 0.75
3
Subtract the indices: $1 0^{12 - 6} = 1 0^{6}$ 10¹²⁻⁶ = 10⁶, giving $0.75 \times 1 0^{6}$ 0.75 × 10⁶
4
Adjust so $1 \leq a < 10$ 1 ≤ a < 10: $0.75 \times 1 0^{6} = 7.5 \times 1 0^{5}$ 0.75 × 10⁶ = 7.5 × 10⁵

Answer

7.5 \times 1 0^{5}

photos (750,000)

⚠️ Watch out

0.75 \times 1 0^{6}

is the right value but not standard form. Making

a

ten times bigger makes the power one smaller.

Calculate

?Write

(2.5 \times 1 0^{- 4}) \times (8 \times 1 0^{9})

in standard form,

a \times 1 0^{k}

. Enter the value of

k

Calculate

?A strand of hair is

8.4 \times 1 0^{- 5}

m wide. A virus particle is

1.2 \times 1 0^{- 7}

m wide. How many virus particles, side by side, would span the width of the hair? Give an exact answer.

Mission check

?A cell is about

1 0^{- 5}

m wide and the Milky Way about

1 0^{21}

m across. How many cells, end to end, would cross it?

Mission 2

20 seats in row 1, two more in each row: how many seats in 30 rows?

Make a guess: 20 seats in row 1, two more in each row: how many seats in 30 rows?

Number and algebra · arithmetic sequences and series

Arithmetic sequences

An arithmetic sequence adds the same amount each time: the common difference $d = u_{n + 1} - u_{n}$ d = uₙ₊₁ − uₙ.
With first term $u_{1}$ u₁, the $n$ nth term is

u_{n} = u_{1} + (n - 1) d

It is $(n - 1) d$ (n−1)d, not $n d$ nd: from $u_{1}$ u₁ to $u_{n}$ uₙ there are $n - 1$ n − 1 steps.
Plotted against $n$ n, the terms lie on a straight line with gradient $d$ d.

Worked example Number and algebra · arithmetic sequences and series

Finding a term and a term number

Question

An arithmetic sequence begins 7, 11, 15, … (a) Find

u_{40}

. (b) Which term is equal to 203?

1
Identify the first term and common difference: $u_{1} = 7$ u₁ = 7, $d = 11 - 7 = 4$ d = 11 − 7 = 4
2
(a) Substitute $n = 40$ n = 40: $u_{40} = 7 + 39 \times 4 = 163$ u₄₀ = 7 + 39 × 4 = 163
3
(b) Set $u_{n} = 203$ uₙ = 203: $7 + (n - 1) \times 4 = 203$ 7 + (n−1) × 4 = 203
4
Solve: $(n - 1) \times 4 = 196$ (n−1) × 4 = 196, so $n - 1 = 49$ n − 1 = 49 and $n = 50$ n = 50

Answer

(a)

u_{40} = 163

(b) 203 is the 50th term

⚠️ Watch out

in (b) the answer must be a positive integer. A decimal value of

n

would mean the number is not a term of the sequence.

Quick check

?An arithmetic sequence has

u_{1} = 5

and

d = - 3

. Find

u_{20}

Number and algebra · arithmetic sequences and series

Arithmetic series and sigma notation

An arithmetic series is the sum of the terms. The sum of the first $n$ n terms is

S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d) = \frac{n}{2} (u_{1} + u_{n})

The second form pairs first and last terms: each pair has the same total.
Sigma notation: $r = 1 \sum n (3 r + 2)$ Σ_(r=1)ⁿ (3r + 2) means "substitute $r = 1, 2, \dots, n$ r = 1, 2, …, n and add".
A GDC can generate terms and sums, but you must still state $u_{1}$ u₁ and $d$ d.

Worked example Number and algebra · arithmetic sequences and series

Evaluating a sum in sigma notation

Question

Find

r = 1 \sum 25 (3 r + 2)

1
Write the first terms: $r = 1$ r = 1 gives 5, $r = 2$ r = 2 gives 8, so $u_{1} = 5$ u₁ = 5 and $d = 3$ d = 3
2
Find the last term, $r = 25$ r = 25: $u_{25} = 3 \times 25 + 2 = 77$ u₂₅ = 3 × 25 + 2 = 77
3
Use $S_{n} = \frac{n}{2} (u_{1} + u_{n})$ Sₙ = n/2(u₁ + uₙ) with $n = 25$ n = 25: $S_{25} = \frac{25}{2} (5 + 77)$ S₂₅ = 25/2(5 + 77)
4
Evaluate: $12.5 \times 82 = 1025$ 12.5 × 82 = 1025

Answer

1025

⚠️ Watch out

d

is the coefficient of

r

(here 3), but

u_{1}

is not the constant 2. Substitute

r = 1

to get

u_{1} = 5

Calculate

?Ana invests 2,000 euros at 3.5% simple interest per year. Simple interest adds 3.5% of the original amount each year. After how many complete years will her investment first be worth more than 3,000 euros?

Number and algebra · arithmetic sequences and series

When data are nearly arithmetic

Real data rarely go up by exactly the same amount. If the differences are roughly equal, an arithmetic model is reasonable.
Approximate the common difference, for example by the mean difference: $d \approx \frac{last value - first value}{number of steps}$ d ≈ (last value − first value)/(number of steps).
Use the model to predict, then comment: predictions far beyond the data are less reliable.

Calculate

?Rio records the distance of his longest run each week: 5.0, 5.9, 7.1, 7.9, 9.0 km for weeks 1 to 5. Use an arithmetic model with first term 5.0 km and the mean difference of the data to predict his longest run in week 12. Give your answer to 3 significant figures.

Calculate

?In an arithmetic sequence,

u_{3} = 17

and

u_{10} = 52

. Find the sum of the first 20 terms.

Mission check

?Row 1 has 20 seats and each row has 2 more than the one before. How many seats are in 30 rows?

Mission 3

Double the rice on each chessboard square. How much on all 64?

Make a guess: Double the rice on each chessboard square. How much on all 64?

Number and algebra · geometric sequences and series

Geometric sequences

A geometric sequence multiplies by the same number each time: the common ratio $r = \frac{u _{n + 1}}{u _{n}}$ r = (uₙ₊₁)/uₙ.
With first term $u_{1}$ u₁, the $n$ nth term is

u_{n} = u_{1} r^{n - 1}

Again $n - 1$ n − 1 steps: $u_{1}$ u₁ is multiplied by $r$ r a total of $n - 1$ n − 1 times.
$r > 1$ r > 1: growth. $0 < r < 1$ 0 < r < 1: decay. $r < 0$ r < 0: the signs alternate.

Worked example Number and algebra · geometric sequences and series

A decreasing geometric sequence

Question

A geometric sequence begins 80, 60, 45, … Find

u_{8}

. Give your answer to 3 significant figures.

1
Find the common ratio: $r = \frac{60}{80} = 0.75$ r = 60/80 = 0.75 (check: $\frac{45}{60} = 0.75$ 45/60 = 0.75)
2
Use $u_{n} = u_{1} r^{n - 1}$ uₙ = u₁ rⁿ⁻¹ with $n = 8$ n = 8: $u_{8} = 80 \times 0.7 5^{7}$ u₈ = 80 × 0.75⁷
3
Evaluate: $80 \times 0.133 \dots = 10.678 \dots$ 80 × 0.133… = 10.678…

Answer

u_{8} = 10.7

(3 sf)

⚠️ Watch out

r

is next term divided by previous term. Using

\frac{80}{60}

gives a ratio above 1 for a sequence that is shrinking.

Quick check

?A geometric sequence begins 2, −6, 18, … Find

u_{6}

Number and algebra · geometric sequences and series

Geometric series

The sum of the first $n$ n terms, for $r \neq = 1$ r ≠ 1:

S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}

Either form works; the first is tidier when $r > 1$ r > 1.
Sigma notation: $k = 1 \sum n u_{1} r^{k - 1}$ Σ_(k=1)ⁿ u₁ r^(k−1) adds the first $n$ n terms.
Applications: salaries rising by a fixed percentage, populations, the spread of a disease. With a GDC, still state $u_{1}$ u₁ and $r$ r.

Calculate

?Mai's starting salary is 32,000 dollars a year. It rises by 3% each year. Find the total she earns in her first 10 years. Give your answer to the nearest dollar.

dollars

Calculate

?A geometric sequence has

u_{1} = 5

and

r = 2

. Find the first term that is greater than 10,000. Enter its term number

n

Number and algebra · geometric sequences and series

Sum to infinity

If $∣ r ∣ < 1$ |r| < 1, the terms shrink towards 0 and $S_{n}$ Sₙ gets closer and closer to a limit. The series converges.
The sum to infinity is

S_{\infty} = \frac{u _{1}}{1 - r}, ∣ r ∣ < 1

If $∣ r ∣ \geq 1$ |r| ≥ 1 the series has no sum to infinity: it diverges.
Example: $1 + \frac{1}{2} + \frac{1}{4} + \dots = \frac{1}{1 - 0.5} = 2$ 1 + 1/2 + 1/4 + ⋯ = 1/(1 − 0.5) = 2.

Worked example Number and algebra · geometric sequences and series

A bouncing ball

Question

A ball is dropped from a height of 2 m. After each bounce it rises to 70% of the height it fell from. Find the total distance it travels before it comes to rest.

1
First drop: 2 m. After that, each bounce goes up and comes down the same height
2
Bounce heights: $1.4, 0.98, \dots$ 1.4, 0.98, …, geometric with $u_{1} = 1.4$ u₁ = 1.4 and $r = 0.7$ r = 0.7, and $∣ r ∣ < 1$ |r| < 1
3
Sum of the bounce heights: $S_{\infty} = \frac{1.4}{1 - 0.7} = 4.666 \dots$ S_(∞) = 1.4/(1 − 0.7) = 4.666… m
4
Total: $2 + 2 \times 4.666 \dots = 11.333 \dots$ 2 + 2 × 4.666… = 11.333… m

Answer

11.3

m (3 sf), exactly

\frac{34}{3}

⚠️ Watch out

the first drop is only down. Every later height is travelled twice, up and down.

Calculate

?A geometric series has first term 12 and sum to infinity 30. Find the common ratio

r

. Give an exact answer as a decimal.

Mission check

?One grain on the first square, then double on each square. How many grains on all 64 squares?

Mission 4

£10,000 at 4% for 10 years: what does monthly compounding add?

Make a guess: £10,000 at 4% for 10 years: what does monthly compounding add?

Number and algebra · financial mathematics

Compound interest

Simple interest adds the same amount every year (an arithmetic sequence). Compound interest earns interest on the interest (a geometric sequence).
With present value $P V$ PV, nominal annual rate $r %$ r%, compounded $k$ k times a year for $n$ n years, the future value is

F V = P V \times (1 + \frac{r}{100 k})^{kn}

Yearly: $k = 1$ k = 1. Half-yearly: $k = 2$ k = 2. Quarterly: $k = 4$ k = 4. Monthly: $k = 12$ k = 12.
A GDC finance solver gives the same answers.

Worked example Number and algebra · financial mathematics

Compounding quarterly

Question

5,000 euros is invested at 3.6% a year, compounded quarterly. Find its value after 6 years. Give your answer to 2 decimal places.

1
Identify the values: $P V = 5000$ PV = 5000, $r = 3.6$ r = 3.6, $k = 4$ k = 4, $n = 6$ n = 6
2
Rate per quarter: $\frac{3.6}{100 \times 4} = 0.009$ 3.6/(100 × 4) = 0.009, so each quarter multiplies by $1.009$ 1.009
3
Number of quarters: $kn = 4 \times 6 = 24$ kn = 4 × 6 = 24
4
Evaluate: $F V = 5000 \times 1.00 9^{24} = 6199.518 \dots$ FV = 5000 × 1.009²⁴ = 6199.518…

Answer

6,199.52 euros

⚠️ Watch out

both the rate and the power change.

5000 \times 1.03 6^{24}

uses the yearly rate every quarter.

Quick check

?2,000 dollars is invested at 6% a year, compounded monthly, for 5 years. Which expression gives its final value?

Number and algebra · financial mathematics

Depreciation, inflation and time

Depreciation: a value falls by $r %$ r% each year, so $F V = P V \times (1 - \frac{r}{100})^{n}$ FV = PV × (1 − r/100)ⁿ.
Real value: inflation reduces what money can buy. IB markschemes use the real interest rate ≈ interest rate − inflation rate in the compound interest formula. This is an approximation: dividing by the inflation factor gives slightly less.
To find how long, solve for $n$ n with a table of values, the finance solver or logarithms, then round up to a whole period if the question asks when a value is first reached.

Calculate

?A car is bought for 24,000 dollars. Its value depreciates by 15% each year. Find its value after 4 years. Give your answer to 2 decimal places.

dollars

Calculate

?8,000 pounds is invested at 5% a year, compounded annually. After how many complete years will it first be worth more than 12,000 pounds?

Calculate

?Zoe invests 10,000 dollars for 8 years at 4.5% a year, compounded annually. Inflation is 2.5% a year. Using the rate (interest rate − inflation rate), find the real value of her investment after 8 years. Give your answer to the nearest dollar.

dollars

Mission check

?£10,000 earns 4% a year for 10 years. How much more does monthly compounding give than yearly?

Mission 5

A magnitude 8 earthquake versus a magnitude 6: how much bigger?

Make a guess: A magnitude 8 earthquake versus a magnitude 6: how much bigger?

Number and algebra · exponents and logarithms

Laws of exponents

$a^{m} \times a^{n} = a^{m + n}$ a^m × aⁿ = a^(m+n), $a^{m} \div a^{n} = a^{m - n}$ a^m ÷ aⁿ = a^(m−n), $(a^{m})^{n} = a^{mn}$ (a^m)ⁿ = a^mn
$a^{0} = 1$ a⁰ = 1 and $a^{- n} = \frac{1}{a ^{n}}$ a⁻ⁿ = 1/aⁿ. So $5^{3} \times 5^{- 6} = 5^{- 3}$ 5³ × 5⁻⁶ = 5⁻³.
A power applies to every factor in its bracket: $(2 x)^{4} = 16 x^{4}$ (2x)⁴ = 16x⁴. But $2 x^{- 3} = \frac{2}{x ^{3}}$ 2x⁻³ = 2/x³: only $x$ x is raised to $- 3$ −3.
Rational exponents: $a^{\frac{1}{m}} = m a$ a^(1/m) = m√a and $a^{\frac{m}{n}} = (n a)^{m}$ a^(m/n) = (n√a)^m, so $1 6^{\frac{3}{4}} = 2^{3} = 8$ 16^(3/4) = 2³ = 8.

Quick check

?Simplify

\frac{( 3 x ^{2} ) ^{3}}{9 x ^{- 1}}

Number and algebra · exponents and logarithms

Logarithms

A logarithm is an index: $a^{x} = b ⟺ lo g_{a} b = x$ a^x = b ⇔ log_a b = x, for $a > 0$ a > 0, $a \neq = 1$ a ≠ 1 and $b > 0$ b > 0.
$lo g_{10}$ log₁₀ is written $lo g$ log; $lo g_{e}$ log_e is written $ln$ ln. Your GDC evaluates both.
$lo g_{a} a = 1$ log_a a = 1 and $lo g_{a} 1 = 0$ log_a 1 = 0. There is no logarithm of 0 or of a negative number.
On a base-10 log scale, each step of 1 is a factor of 10: $lo g 30 = 1.477 \dots$ log 30 = 1.477… and $lo g 300 = 2.477 \dots$ log 300 = 2.477…
The earthquake magnitude scale is such a logarithmic scale.

Worked example Number and algebra · exponents and logarithms

Solving with $lo g$ log and $ln$ ln

Question

Solve (a)

1 0^{x} = 250

(b)

e^{x} = 7

. Give your answers to 3 significant figures.

1
(a) Rewrite as a logarithm: $x = lo g_{10} 250 = lo g 250$ x = log₁₀ 250 = log 250
2
Evaluate with your GDC: $x = 2.397 \dots$ x = 2.397…
3
(b) Rewrite with base e: $x = lo g_{e} 7 = ln 7$ x = log_e 7 = ln 7
4
Evaluate: $x = 1.945 \dots$ x = 1.945…

Answer

(a)

x = 2.40

(b)

x = 1.95

⚠️ Watch out

1 0^{2.40}

is about 250, so the answer is sensible. Check by substituting back.

Calculate

?The pH of a solution is

pH = - lo g_{10} c

, where

c

is the hydrogen ion concentration in mol dm⁻³. Find the pH when

c = 3.2 \times 1 0^{- 4}

. Give your answer to 3 significant figures.

Calculate

?A bacteria colony is modelled by

N = 500 \times 2^{t}

, where

t

is the time in hours. Find the time when

N = 20000

. Use logarithms or your GDC. Give your answer to 3 significant figures.

hours

Number and algebra · exponents and logarithms

Laws of logarithms

For $a, x, y > 0$ a, x, y > 0:

lo g_{a} x y = lo g_{a} x + lo g_{a} y, lo g_{a} \frac{x}{y} = lo g_{a} x - lo g_{a} y, lo g_{a} x^{m} = m lo g_{a} x

Example: $lo g 24 = lo g 8 + lo g 3$ log 24 = log 8 + log 3, and $lo g 32 = 5 lo g 2$ log 32 = 5 log 2.
The power law brings an unknown index down: take $ln$ ln (or $lo g$ log) of both sides.
Change of base: $lo g_{a} x = \frac{lo g _{b} x}{lo g _{b} a}$ log_a x = (log_b x)/(log_b a), e.g. $lo g_{4} 7 = \frac{ln 7}{ln 4}$ log₄ 7 = (ln 7)/(ln 4).

Worked example Number and algebra · exponents and logarithms

An exponential equation

Question

Solve

3^{2 x - 1} = 20

. Give your answer to 3 significant figures.

1
Take $ln$ ln of both sides: $ln (3^{2 x - 1}) = ln 20$ ln (3^(2x−1)) = ln 20
2
Use the power law: $(2 x - 1) ln 3 = ln 20$ (2x − 1)ln 3 = ln 20
3
Divide by $ln 3$ ln 3: $2 x - 1 = \frac{ln 20}{ln 3} = 2.726 \dots$ 2x − 1 = (ln 20)/(ln 3) = 2.726…
4
Solve: $x = \frac{2.726 \dots + 1}{2} = 1.863 \dots$ x = (2.726… + 1)/2 = 1.863…

Answer

x = 1.86

⚠️ Watch out

\frac{l n 20}{l n 3}

is not

ln \frac{20}{3}

. Dividing two logarithms is not the quotient law.

Calculate

?Given that

2 ln 6 - ln 4 = ln k

, find the value of

k

Mission check

?The magnitude scale is logarithmic (base 10). How many times larger is the measured amplitude of a magnitude 8 earthquake than a magnitude 6?

Mission 6

Why do the digits of $1 1^{4} = 14641$ 11⁴ = 14641 form a row of Pascal's triangle?

Make a guess: Why do the digits of $1 1^{4} = 14641$ 11⁴ = 14641 form a row of Pascal's triangle?

Number and algebra · proof and the binomial theorem

Identities and LHS to RHS proof

An identity is true for every value of the variable; write it with $\equiv$ ≡. An equation, with $=$ =, is true only for some values.
To show that LHS $\equiv$ ≡ RHS, start with one side and transform it, one valid algebraic step per line, until you reach the other side. Never work on both sides at once.
Check your result by substituting a value. If $\frac{1}{m + 1} + \frac{1}{m ^{2} + m} \equiv \frac{1}{m}$ 1/(m+1) + 1/(m²+m) ≡ 1/m, then $m = 3$ m = 3 must give $\frac{1}{4} + \frac{1}{12} = \frac{1}{3}$ 1/4 + 1/12 = 1/3, which it does.

Worked example Number and algebra · proof and the binomial theorem

A deductive proof from left to right

Question

Show that

\frac{1}{m + 1} + \frac{1}{m ^{2} + m} \equiv \frac{1}{m}

1
Factorise the second denominator on the LHS: $m^{2} + m = m (m + 1)$ m² + m = m(m+1)
2
Write both fractions over $m (m + 1)$ m(m+1): LHS $= \frac{m}{m ( m + 1 )} + \frac{1}{m ( m + 1 )}$ = m/(m(m+1)) + 1/(m(m+1))
3
Add the numerators: $= \frac{m + 1}{m ( m + 1 )}$ = (m+1)/(m(m+1))
4
Cancel the common factor $m + 1$ m+1: $= \frac{1}{m} =$ = 1/m = RHS

Answer

LHS

\equiv

RHS, as required.

⚠️ Watch out

writing the whole identity on every line and "doing the same to both sides" assumes what you are trying to prove. Start from one side only.

Sort it

?Is each statement an identity (true for every

x

) or an equation (true only for some

x

Tap a statement, then tap the group it belongs in. Then press Check.

Number and algebra · proof and the binomial theorem

The binomial theorem and Pascal's triangle

Each row of Pascal's triangle starts and ends with 1; every other entry is the sum of the two above it. Row $n$ n gives the coefficients of $(a + b)^{n}$ (a+b)ⁿ.
The entries are $(r n) = \frac{n !}{r ! ( n - r )!}$ C(n, r) = (n!)/(r!(n−r)!), also written $^{n} C_{r}$ ⁿCᵣ. Find them by the formula or with your GDC.
Binomial theorem for $n \in N$ n ∈ ℕ:

(a + b)^{n} = a^{n} + (1 n) a^{n - 1} b + \dots + (r n) a^{n - r} b^{r} + \dots + b^{n}

The general term is $(r n) a^{n - r} b^{r}$ C(n, r)a^(n−r)b^r. The powers of $a$ a and $b$ b always add to $n$ n.

Calculate

?Use your GDC to make a table of values of

(r 6)

for

r = 0, 1, \dots, 6

. Find the value of

r

for which

(r 6) = 20

Worked example Number and algebra · proof and the binomial theorem

One coefficient from the general term

Question

Find the coefficient of

x^{3}

in the expansion of

(2 x - 3)^{5}

1
Write the general term with $a = 2 x$ a = 2x and $b = - 3$ b = −3: $(r 5) (2 x)^{5 - r} (- 3)^{r}$ C(5, r)(2x)^(5−r)(−3)^r
2
For $x^{3}$ x³ you need $5 - r = 3$ 5 − r = 3, so $r = 2$ r = 2
3
Evaluate each part: $(2 5) = 10$ C(5, 2) = 10, $(2 x)^{3} = 8 x^{3}$ (2x)³ = 8x³, $(- 3)^{2} = 9$ (−3)² = 9
4
Multiply: $10 \times 8 \times 9 = 720$ 10 × 8 × 9 = 720

Answer

720

⚠️ Watch out

the power applies to the whole bracket. Writing

2 x^{3}

instead of

(2 x)^{3} = 8 x^{3}

gives 180, and dropping the bracket on

- 3

gets the sign wrong.

Calculate

?Find the constant term in the expansion of

(x^{2} + \frac{2}{x})^{6}

Calculate

?In the expansion of

(1 + k x)^{5}

, where

k > 0

, the coefficient of

x^{2}

is 40. Find

k

Quick check

?Find the first three terms, in ascending powers of

x

, of

(1 - \frac{x}{2})^{8}

Mission check

?Why do the digits of

1 1^{4} = 14641

form a row of Pascal's triangle?

Mission 7

How many ways can 8 runners finish 1st, 2nd and 3rd?

Make a guess: How many ways can 8 runners finish 1st, 2nd and 3rd?

Number and algebra · counting and the binomial theorem

Counting principles: permutations and combinations

Multiplication principle: if one choice can be made in $p$ p ways and the next in $q$ q ways, together they can be made in $p \times q$ p × q ways. 5 runners on a podium: $5 \times 4 \times 3 = 60$ 5 × 4 × 3 = 60 ways.
$n$ n objects in a row: $n! = n (n - 1) \dots 2 \times 1$ n! = n(n−1)⋯ 2 × 1 ways.
Permutation (order matters): $^{n} P_{r} = \frac{n !}{( n - r )!}$ ⁿPᵣ = (n!)/((n−r)!) ways to choose and arrange $r$ r of $n$ n.
Combination (order does not matter): $^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}$ ⁿCᵣ = C(n, r) = (n!)/(r!(n−r)!). Each choice of $r$ r objects can be ordered in $r!$ r! ways, so $^{n} C_{r} =^{n} P_{r} \div r!$ ⁿCᵣ =ⁿPᵣ ÷ r!.

Worked example Number and algebra · counting and the binomial theorem

A selection with a condition

Question

A team of 5 is chosen from 6 girls and 4 boys. Find the number of teams that contain at least 3 girls.

1
Split "at least 3 girls" into cases: 3 girls and 2 boys, 4 girls and 1 boy, 5 girls and 0 boys
2
3 girls, 2 boys: $(3 6) (2 4) = 20 \times 6 = 120$ C(6, 3)C(4, 2) = 20 × 6 = 120
3
4 girls, 1 boy: $(4 6) (1 4) = 15 \times 4 = 60$ C(6, 4)C(4, 1) = 15 × 4 = 60
4
5 girls: $(5 6) (0 4) = 6$ C(6, 5)C(4, 0) = 6
5
The cases cannot happen together, so add: $120 + 60 + 6 = 186$ 120 + 60 + 6 = 186

Answer

186 teams

⚠️ Watch out

multiply within a case (girls and boys), add across cases (this case or that one). Choosing 3 girls first and then any 2 of the other 7 people counts some teams more than once.

Sort it

?Does order matter? Classify each count.

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?Seven friends sit in a row of 7 seats for a photo. Two of them, Amal and Bea, must sit next to each other. Find the number of possible arrangements.

Number and algebra · counting and the binomial theorem

The binomial theorem for any rational index

For $n \in Q$ n ∈ ℚ (including negative and fractional $n$ n):

(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots

If $n$ n is not a positive integer, the series never ends, and it is valid only for $∣ x ∣ < 1$ |x| < 1.
For $(a + b)^{n}$ (a+b)ⁿ, take out $a$ a first: $(a + b)^{n} = a^{n} (1 + \frac{b}{a})^{n}$ (a+b)ⁿ = aⁿ(1 + b/a)ⁿ, valid for $\frac{b}{a} < 1$ |b/a| < 1.
Replace $x$ x by the whole term, sign and coefficient included, e.g. $(1 - 3 x)^{- 2}$ (1−3x)⁻² uses $- 3 x$ −3x.

Worked example Number and algebra · counting and the binomial theorem

A fractional index and an estimate

Question

Expand

4 + x

in ascending powers of

x

up to

x^{2}

, state the values of

x

for which it is valid, and hence estimate

4.2

1
Take out the 4: $(4 + x)^{\frac{1}{2}} = 4^{\frac{1}{2}} (1 + \frac{x}{4})^{\frac{1}{2}} = 2 (1 + \frac{x}{4})^{\frac{1}{2}}$ (4+x)^(1/2) = 4^(1/2)(1 + x/4)^(1/2) = 2(1 + x/4)^(1/2)
2
Use $n = \frac{1}{2}$ n = 1/2 with $\frac{x}{4}$ x/4: $1 + \frac{1}{2} (\frac{x}{4}) + \frac{\frac{1}{2} ( - \frac{1}{2} )}{2} (\frac{x}{4})^{2} = 1 + \frac{x}{8} - \frac{x ^{2}}{128}$ 1 + 1/2(x/4) + (1/2(−1/2))/2(x/4)² = 1 + x/8 − x²/128
3
Multiply by 2: $4 + x \approx 2 + \frac{x}{4} - \frac{x ^{2}}{64}$ √(4+x) ≈ 2 + x/4 − x²/64
4
Validity: $\frac{x}{4} < 1$ |x/4| < 1, so $∣ x ∣ < 4$ |x| < 4
5
Put $x = 0.2$ x = 0.2: $2 + 0.05 - 0.000625 = 2.049375$ 2 + 0.05 − 0.000625 = 2.049375

Answer

2 + \frac{x}{4} - \frac{x ^{2}}{64}

for

∣ x ∣ < 4

;

4.2 \approx 2.049

(a GDC gives 2.04939…)

⚠️ Watch out

forgetting to take

4^{\frac{1}{2}} = 2

outside, or writing

4^{\frac{1}{2}}

as 4, puts every term out by a factor.

Quick check

?For which values of

x

is the expansion of

(2 - 3 x)^{- 2}

in ascending powers of

x

valid?

Calculate

?Find the coefficient of

x^{2}

in the expansion of

(1 + 2 x)^{- 3}

Mission check

?How many ways can 8 runners finish 1st, 2nd and 3rd?

Mission 8

Three equations, three unknowns: how can there be no solution?

Make a guess: Three equations, three unknowns: how can there be no solution?

Number and algebra · partial fractions and linear systems

Partial fractions

A fraction with two distinct linear factors in the denominator can be split into partial fractions:

\frac{p x + q}{( x - a ) ( x - b )} \equiv \frac{A}{x - a} + \frac{B}{x - b}

This works when the degree of the numerator is less than the degree of the denominator. Factorise the denominator first.
Multiply through by $(x - a) (x - b)$ (x−a)(x−b) to get $p x + q \equiv A (x - b) + B (x - a)$ px + q ≡ A(x−b) + B(x−a). It is an identity, so substitute $x = a$ x = a and $x = b$ x = b, or equate coefficients.

Worked example Number and algebra · partial fractions and linear systems

Splitting into partial fractions

Question

Express

\frac{5 x + 1}{x ^{2} + x - 2}

in partial fractions.

1
Factorise the denominator: $x^{2} + x - 2 = (x - 1) (x + 2)$ x² + x − 2 = (x−1)(x+2)
2
Write the form and clear fractions: $5 x + 1 \equiv A (x + 2) + B (x - 1)$ 5x + 1 ≡ A(x+2) + B(x−1)
3
Put $x = 1$ x = 1: $6 = 3 A$ 6 = 3A, so $A = 2$ A = 2
4
Put $x = - 2$ x = −2: $- 9 = - 3 B$ −9 = −3B, so $B = 3$ B = 3

Answer

\frac{5 x + 1}{x ^{2} + x - 2} \equiv \frac{2}{x - 1} + \frac{3}{x + 2}

⚠️ Watch out

check by recombining:

2 (x + 2) + 3 (x - 1) = 5 x + 1

. A sign slip in the factors, such as using

(x + 1) (x - 2)

, makes the check fail.

Calculate

?Given that

\frac{3 x + 11}{( x - 3 ) ( x + 1 )} \equiv \frac{A}{x - 3} + \frac{B}{x + 1}

, find the value of

B

Number and algebra · partial fractions and linear systems

Systems of three linear equations

Each equation $a x + b y + cz = d$ ax + by + cz = d is a plane. A solution is a point on all three planes.
Solve by elimination (row reduction): use one equation to remove a variable from the other two, then solve the pair. Or use your GDC's solver or matrices.
Three outcomes: a unique solution (one common point); infinitely many solutions, when elimination gives $0 = 0$ 0 = 0 (the planes share a line); no solution, when elimination gives a false statement such as $0 = 5$ 0 = 5. A system with no solution is inconsistent.

Worked example Number and algebra · partial fractions and linear systems

A system with a unique solution

Question

Solve

x + y + z = 6

2 x - y + z = 3

x + 2 y - z = 2

1
Add the first and third equations to remove $z$ z: $2 x + 3 y = 8$ 2x + 3y = 8
2
Add the second and third equations to remove $z$ z: $3 x + y = 5$ 3x + y = 5
3
From the second result, $y = 5 - 3 x$ y = 5 − 3x. Substitute: $2 x + 15 - 9 x = 8$ 2x + 15 − 9x = 8, so $x = 1$ x = 1
4
Back-substitute: $y = 5 - 3 = 2$ y = 5 − 3 = 2, then $z = 6 - 1 - 2 = 3$ z = 6 − 1 − 2 = 3

Answer

x = 1

y = 2

z = 3

⚠️ Watch out

always check in the equation you did not use last:

2 (1) - 2 + 3 = 3

. ✓

Calculate

?A cinema sells adult, child and senior tickets at prices

a

c

and

s

euros. One group pays 54 euros for 2 adult, 3 child and 1 senior ticket. A second pays 44 euros for 1 adult, 2 child and 2 senior tickets. A third pays 61 euros for 3 adult, 1 child and 2 senior tickets. Use your GDC to find the price of one senior ticket.

Worked example Number and algebra · partial fractions and linear systems

Infinitely many solutions: the general solution

Question

Show that the system

x + y + z = 2

x + 2 y + 3 z = 5

2 x + 3 y + 4 z = 7

has infinitely many solutions, and find the general solution.

1
Second minus first: $y + 2 z = 3$ y + 2z = 3
2
Third minus twice the first: $y + 2 z = 3$ y + 2z = 3
3
Subtracting these gives $0 = 0$ 0 = 0, so only two independent equations remain: infinitely many solutions
4
Let $z = λ$ z = λ. Then $y = 3 - 2 λ$ y = 3 − 2λ and $x = 2 - y - z = - 1 + λ$ x = 2 − y − z = −1 + λ

Answer

x = - 1 + λ

y = 3 - 2 λ

z = λ

, for

λ \in R

⚠️ Watch out

many GDC solvers just report "infinitely many solutions" or an error here. You still have to find the general solution by elimination.

Quick check

?The third equation in the last example is changed to

2 x + 3 y + 4 z = 10

. What happens to the system

x + y + z = 2

x + 2 y + 3 z = 5

2 x + 3 y + 4 z = 10

Mission check

?Three equations, three unknowns: how can there be no solution?

Recap

Big ideas

Standard form is $a \times 1 0^{k}$ a × 10^k with $1 \leq a < 10$ 1 ≤ a < 10 and $k$ k an integer; never write calculator notation like 5.2E30.
To multiply, add the indices; to divide, subtract them; then adjust so $1 \leq a < 10$ 1 ≤ a < 10.
Arithmetic sequence: $u_{n} = u_{1} + (n - 1) d$ uₙ = u₁ + (n−1)d, where $d$ d is the common difference.
Arithmetic series: $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d) = \frac{n}{2} (u_{1} + u_{n})$ Sₙ = n/2(2u₁ + (n−1)d) = n/2(u₁ + uₙ); sigma notation lists the terms to add.
Simple interest grows arithmetically; for real data, approximate $d$ d and judge how far a prediction can be trusted.
Geometric sequence: $u_{n} = u_{1} r^{n - 1}$ uₙ = u₁ rⁿ⁻¹, where $r$ r is the common ratio (next term ÷ previous term).
Geometric series: $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ Sₙ = (u₁(rⁿ − 1))/(r − 1) = (u₁(1 − rⁿ))/(1 − r), $r \neq = 1$ r ≠ 1.
If $∣ r ∣ < 1$ |r| < 1 the series converges and $S_{\infty} = \frac{u _{1}}{1 - r}$ S_(∞) = u₁/(1 − r).
Compound interest: $F V = P V (1 + \frac{r}{100 k})^{kn}$ FV = PV(1 + r/100k)^kn, with $k$ k compounding periods a year; depreciation multiplies by $1 - \frac{r}{100}$ 1 − r/100 each year.
For real value, use the interest rate minus the inflation rate (the approximation IB uses); to find a time, solve for $n$ n and round up to a whole period.
Laws of exponents: add indices to multiply, subtract to divide, multiply for a power; $a^{- n} = \frac{1}{a ^{n}}$ a⁻ⁿ = 1/aⁿ.
$a^{x} = b ⟺ lo g_{a} b = x$ a^x = b ⇔ log_a b = x; $lo g$ log means base 10 and $ln$ ln means base e. On a base-10 log scale, +1 means ×10.
Laws of logarithms turn products into sums and powers into multipliers; take logs to solve $a^{f (x)} = b$ a^(f(x)) = b.
Show that LHS $\equiv$ ≡ RHS by starting from one side and transforming it step by step; check with a value.
$(a + b)^{n} = \sum_{r = 0 n} (r n) a^{n - r} b^{r}$ (a+b)ⁿ = Σ_(r=0)ⁿC(n, r)a^(n−r)b^r for $n \in N$ n ∈ ℕ, with $(r n) = \frac{n !}{r ! ( n - r )!}$ C(n, r) = (n!)/(r!(n−r)!) from Pascal's triangle, the formula or a GDC.
To find one coefficient, use the general term, choose $r$ r from the power you want, and keep brackets round each part.
Multiply the choices at each stage; order matters gives $^{n} P_{r} = \frac{n !}{( n - r )!}$ ⁿPᵣ = (n!)/((n−r)!), order does not matter gives $(r n) = \frac{n !}{r ! ( n - r )!}$ C(n, r) = (n!)/(r!(n−r)!).
With conditions, split into cases and add, or glue items together into a block.
For $n \in Q$ n ∈ ℚ, $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$ (1+x)ⁿ = 1 + nx + (n(n−1))/(2!)x² + ⋯ is infinite and valid for $∣ x ∣ < 1$ |x| < 1; write $(a + b)^{n} = a^{n} (1 + \frac{b}{a})^{n}$ (a+b)ⁿ = aⁿ(1+b/a)ⁿ, valid for $\frac{b}{a} < 1$ |b/a| < 1.
$\frac{p x + q}{( x - a ) ( x - b )} \equiv \frac{A}{x - a} + \frac{B}{x - b}$ (px+q)/((x−a)(x−b)) ≡ A/(x−a) + B/(x−b): clear fractions, then substitute $x = a$ x = a and $x = b$ x = b.
Solve three linear equations by elimination or with a GDC: one common point gives a unique solution.
Elimination giving $0 = 0$ 0 = 0 means infinitely many solutions (write the general solution with a parameter); a false statement like $0 = 3$ 0 = 3 means no solution (inconsistent).

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Number & Algebra for IB Diploma Mathematics: Analysis & Approaches HL. 🎉

Mission 1How many cells laid end to end would cross the Milky Way?
Mission 220 seats in row 1, two more in each row: how many seats in 30 rows?
Mission 3Double the rice on each chessboard square. How much on all 64?
Mission 4£10,000 at 4% for 10 years: what does monthly compounding add?
Mission 5A magnitude 8 earthquake versus a magnitude 6: how much bigger?
Mission 6Why do the digits of $1 1^{4} = 14641$ 11⁴ = 14641 form a row of Pascal's triangle?
Mission 7How many ways can 8 runners finish 1st, 2nd and 3rd?
Mission 8Three equations, three unknowns: how can there be no solution?

Your stars: 0 / 0

Next: more IB Mathematics: Analysis & Approaches · HL lessons and games on the IB Mathematics: Analysis & Approaches · HL page.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to IB Mathematics: Analysis & Approaches · HL

Specification points and sources for this lesson are listed at the bottom of the page.

Sources (12)

The specification points and sources behind this lesson.

1 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 1.1; prior learning list
2 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 1.2
3 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 1.2 connections (chess legend), SL 1.3, SL 1.8
4 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 1.4
5 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 1.5, SL 1.7
6 IB Mathematics: analysis and approaches guide (first assessment 2021), SL 1.6 and SL 1.9 content and guidance
7 IB Mathematics: analysis and approaches guide (first assessment 2021), AHL 1.10 content and guidance
8 IB Mathematics: analysis and approaches guide (first assessment 2021), AHL 1.11 and AHL 1.16 content and guidance
9 AAT Bioquest, "What are the size ranges of mammalian cells?" (most mammalian cells are 10–100 µm in diameter), https://aatbio.com/resources/faq-frequently-asked-questions/What-are-the-size-ranges-of-mammalian-cells
10 NASA Imagine the Universe, "The Milky Way Galaxy" (approximately 100,000 light-years in diameter), https://imagine.gsfc.nasa.gov/science/featured_science/milkyway
11 IAU definition of the light-year: distance light travels in a Julian year, 9,460,730,472,580.8 km (about $9.46 \times 1 0^{15}$ 9.46 × 10¹⁵ m), IAU "Measuring the Universe", https://iauarchive.eso.org/public/themes/measuring/
12 USGS, "The Severity of an Earthquake" (each whole number increase in magnitude represents a tenfold increase in measured amplitude; about 31 times more energy per step), https://pubs.usgs.gov/gip/earthq4/severitygip.html

Number & Algebra

How many cells laid end to end would cross the Milky Way?

Standard form

Calculating in standard form

Dividing in standard form

20 seats in row 1, two more in each row: how many seats in 30 rows?

Arithmetic sequences

Finding a term and a term number

Arithmetic series and sigma notation

Evaluating a sum in sigma notation

When data are nearly arithmetic

Double the rice on each chessboard square. How much on all 64?

Geometric sequences

A decreasing geometric sequence

Geometric series

Sum to infinity

A bouncing ball

£10,000 at 4% for 10 years: what does monthly compounding add?

Compound interest

Compounding quarterly

Depreciation, inflation and time

A magnitude 8 earthquake versus a magnitude 6: how much bigger?

Laws of exponents

Logarithms

Solving with loglog and lnln

Laws of logarithms

An exponential equation

Why do the digits of 114=1464111⁴ = 14641 form a row of Pascal's triangle?

Identities and LHS to RHS proof

A deductive proof from left to right

The binomial theorem and Pascal's triangle

One coefficient from the general term

How many ways can 8 runners finish 1st, 2nd and 3rd?

Counting principles: permutations and combinations

A selection with a condition

The binomial theorem for any rational index

A fractional index and an estimate

Three equations, three unknowns: how can there be no solution?

Partial fractions

Splitting into partial fractions

Systems of three linear equations

A system with a unique solution

Infinitely many solutions: the general solution

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

Solving with $lo g$ log and $ln$ ln

Why do the digits of $1 1^{4} = 14641$ 11⁴ = 14641 form a row of Pascal's triangle?