Mini-Lesson

Sequences and series

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

Mission 1How can you work out $1.0 1^{10}$ 1.01¹⁰ to 4 decimal places in your head?
Mission 2 $(1 + x)^{- 1} = 1 - x + x^{2} - \dots$ (1+x)⁻¹=1−x+x²−… Why is it nonsense for $x = 2$ x=2?
Mission 3Add every whole number from 1 to 100 in seconds. How?
Mission 4Halve the distance to a wall again and again. Do you ever arrive?
Mission 5 $u_{n + 1} = \frac{1}{1 - u _{n}}$ uₙ₊₁ = 1/(1−uₙ) and $u_{1} = 2$ u₁ = 2. What is $u_{100}$ u₁₀₀?

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

Mission 1

How can you work out $1.0 1^{10}$ 1.01¹⁰ to 4 decimal places in your head?

Make a guess: How can you work out $1.0 1^{10}$ 1.01¹⁰ to 4 decimal places in your head?

Sequences and series · binomial expansion

Pascal's triangle and $(a + b)^{n}$ (a+b)ⁿ

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. Row 4 is 1, 4, 6, 4, 1:

(a + b)^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4}

In each term the power of $a$ a falls by 1 as the power of $b$ b rises by 1; the powers always add to $n$ n.

Sequences and series · binomial expansion

Factorials and $^{n} C_{r}$ ⁿCᵣ

$n!$ n! $= n \times (n - 1) \times \dots \times 2 \times 1$ =n × (n−1) × … × 2 × 1, and $0! = 1$ 0!=1.
The binomial coefficient $(r n) =^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ C(n, r)=ⁿCᵣ=(n!)/(r! (n−r)!) is entry $r$ r of row $n$ n, counting from $r = 0$ r=0. So $^{5} C_{2} = 10$ ⁵C₂=10.
Binomial expansion: $(a + b)^{n} = a^{n} + (1 n) a^{n - 1} b + \dots + (r n) a^{n - r} b^{r} + \dots + b^{n}$ (a+b)ⁿ=aⁿ+C(n, 1)aⁿ⁻¹b+…+C(n, r)a^(n−r)b^r+…+bⁿ.
$^{n} C_{r}$ ⁿCᵣ counts the ways to pick which $r$ r brackets give a $b$ b; the same count gives binomial probabilities: $P (3 heads in 5 tosses) = (3 5) (\frac{1}{2})^{5} = \frac{10}{32}$ P(3 heads in 5 tosses)=C(5, 3)(1/2)⁵=10/32.

Worked example Sequences and series · binomial expansion

Expanding a bracket in full

Question

Expand

(2 - 3 x)^{4}

fully.

1
Take the row 4 coefficients: 1, 4, 6, 4, 1
2
Write each term as coefficient $\times 2^{4 - r} \times (- 3 x)^{r}$ × 2^(4−r) × (−3x)^r: $2^{4} + 4 (2^{3}) (- 3 x) + 6 (2^{2}) (- 3 x)^{2} + 4 (2) (- 3 x)^{3} + (- 3 x)^{4}$ 2⁴+4(2³)(−3x)+6(2²)(−3x)²+4(2)(−3x)³+(−3x)⁴
3
Evaluate, keeping each bracket whole: $16 - 96 x + 6 (4) (9 x^{2}) - 8 (27 x^{3}) + 81 x^{4}$ 16−96x+6(4)(9x²)−8(27x³)+81x⁴
4
Simplify: $16 - 96 x + 216 x^{2} - 216 x^{3} + 81 x^{4}$ 16−96x+216x²−216x³+81x⁴

Answer

(2 - 3 x)^{4} = 16 - 96 x + 216 x^{2} - 216 x^{3} + 81 x^{4}

⚠️ Watch out

raise the whole bracket

(- 3 x)

to the power, so the signs alternate and the 3 is squared, cubed and so on.

Quick check

?Find the coefficient of

x^{3}

in the expansion of

(2 - 3 x)^{6}

.

Calculate

?In the expansion of

(1 + k x)^{8}

, where

k > 0

, the coefficient of

x^{2}

is 112. Find

k

.

Sequences and series · binomial expansion

Using an expansion to approximate

Write the number as a bracket with a small part: $2.01 = 2 + 0.01$ 2.01=2+0.01, $0.97 = 1 - 3 (0.01)$ 0.97=1−3(0.01).
Expand in powers of $x$ x, then substitute the small value. Each extra power of $x$ x makes the terms much smaller, so the first few terms give a good estimate.
Keep adding terms until the next one is too small to change the digits you need.

Worked example Sequences and series · binomial expansion

Estimating a power

Question

Find the first three terms, in ascending powers of

x

, of

(2 + x)^{6}

. Hence estimate

2.0 1^{6}

.

1
Use row 6 (1, 6, 15, …): $2^{6} + 6 (2^{5}) x + 15 (2^{4}) x^{2}$ 2⁶+6(2⁵)x+15(2⁴)x²
2
Evaluate: $64 + 192 x + 240 x^{2}$ 64+192x+240x²
3
Choose $x$ x so that $2 + x = 2.01$ 2+x=2.01: $x = 0.01$ x=0.01
4
Substitute: $64 + 192 (0.01) + 240 (0.0001) = 64 + 1.92 + 0.024$ 64+192(0.01)+240(0.0001)=64+1.92+0.024

Answer

2.0 1^{6} \approx 65.944

(the true value is

65.94416 \dots

; the next term,

160 x^{3} = 0.00016

, accounts for almost all of the error)

⚠️ Watch out

substitute the value of

x

, not the whole small number

2.01

, into the expansion.

Calculate

?Find the first three terms, in ascending powers of

x

, of

(1 - 3 x)^{5}

. Use your answer, with a suitable value of

x

, to estimate

0.9 7^{5}

. Give your estimate to 3 decimal places.

Mission check

?How can you work out

1.0 1^{10}

to 4 decimal places in your head?

Mission 2

$(1 + x)^{- 1} = 1 - x + x^{2} - \dots$ (1+x)⁻¹=1−x+x²−… Why is it nonsense for $x = 2$ x=2?

Make a guess: $(1 + x)^{- 1} = 1 - x + x^{2} - \dots$ (1+x)⁻¹=1−x+x²−… Why is it nonsense for $x = 2$ x=2?

Sequences and series · binomial expansion, rational

n

The expansion for any rational $n$ n (A2 only)

For any rational $n$ 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 a positive integer the series stops. Otherwise it is infinite, and $^{n} C_{r}$ ⁿCᵣ is not used.
It is valid only for $∣ x ∣ < 1$ |x|<1. Outside that the terms grow and the series diverges.
Replace $x$ x by $k x$ kx for $(1 + k x)^{n}$ (1+kx)ⁿ: valid for $∣ k x ∣ < 1$ |kx|<1.

Worked example Sequences and series · binomial expansion, rational

n

A negative power

Question

Expand

(1 + 3 x)^{- 2}

in ascending powers of

x

up to the term in

x^{3}

, and state the values of

x

for which it is valid.

1
Use $n = - 2$ n=−2 with $3 x$ 3x in place of $x$ x: $1 + (- 2) (3 x) + \frac{( - 2 ) ( - 3 )}{2 !} (3 x)^{2} + \frac{( - 2 ) ( - 3 ) ( - 4 )}{3 !} (3 x)^{3}$ 1+(−2)(3x)+((−2)(−3))/(2!)(3x)²+((−2)(−3)(−4))/(3!)(3x)³
2
Simplify each coefficient: $1 - 6 x + 3 (9 x^{2}) - 4 (27 x^{3})$ 1−6x+3(9x²)−4(27x³)
3
Tidy: $1 - 6 x + 27 x^{2} - 108 x^{3}$ 1−6x+27x²−108x³
4
Validity: $∣3 x ∣ < 1$ |3x|<1, so $∣ x ∣ < \frac{1}{3}$ |x|<1/3

Answer

(1 + 3 x)^{- 2} = 1 - 6 x + 27 x^{2} - 108 x^{3} + \dots

, valid for

∣ x ∣ < \frac{1}{3}

⚠️ Watch out

square and cube the whole of

(3 x)

; writing

3 x^{2}

instead of

9 x^{2}

is the most common lost mark.

Quick check

?For which values of

x

is the expansion of

(1 - 4 x)^{\frac{1}{2}}

valid?

Sequences and series · binomial expansion, rational

n

Expanding $(a + b x)^{n}$ (a+bx)ⁿ

The formula needs a 1 in the bracket, so take out $a^{n}$ aⁿ first:

(a + b x)^{n} = a^{n} (1 + \frac{b x}{a})^{n}

The factor outside is $a$ a raised to the power $n$ n: $(4 + x)^{\frac{1}{2}} = 4^{\frac{1}{2}} (1 + \frac{x}{4})^{\frac{1}{2}} = 2 (1 + \frac{x}{4})^{\frac{1}{2}}$ (4+x)^(1/2)=4^(1/2)(1+x/4)^(1/2)=2(1+x/4)^(1/2).
The expansion is valid for $\frac{b x}{a} < 1$ |bx/a|<1, that is $∣ x ∣ < \frac{a}{b}$ |x|<|a/b|.

Worked example Sequences and series · binomial expansion, rational

n

Taking out a factor, then approximating

Question

Find the first three terms of the expansion of

4 + x

, state its validity, and use it to estimate

4.1

.

1
Take out $4^{\frac{1}{2}} = 2$ 4^(1/2)=2: $4 + x = 2 (1 + \frac{x}{4})^{\frac{1}{2}}$ √(4+x)=2(1+x/4)^(1/2)
2
Expand with $n = \frac{1}{2}$ n=1/2: $2 (1 + \frac{1}{2} (\frac{x}{4}) + \frac{\frac{1}{2} ( - \frac{1}{2} )}{2} (\frac{x}{4})^{2} + \dots)$ 2(1+1/2(x/4)+(1/2(−1/2))/2(x/4)²+…), which is $2 (1 + \frac{x}{8} - \frac{x ^{2}}{128} + \dots)$ 2(1+x/8−x²/128+…)
3
Multiply by 2: $2 + \frac{x}{4} - \frac{x ^{2}}{64}$ 2+x/4−x²/64, valid for $\frac{x}{4} < 1$ |x/4|<1, i.e. $∣ x ∣ < 4$ |x|<4
4
Put $x = 0.1$ x=0.1 (inside the interval): $2 + 0.025 - 0.00015625 = 2.02484375$ 2+0.025−0.00015625=2.02484375

Answer

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

for

∣ x ∣ < 4

, and

4.1 \approx 2.02484

(true value

2.024845 \dots

)

⚠️ Watch out

the factor outside is

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

, not 4.

Calculate

?Find the coefficient of

x^{2}

in the expansion of

\frac{1}{( 2 + x ) ^{2}}

in ascending powers of

x

. Give your answer as a decimal.

Sequences and series · binomial expansion, rational

n

Combining with partial fractions

To expand a fraction such as $\frac{4 - 5 x}{( 1 + x ) ( 2 - x )}$ (4−5x)/((1+x)(2−x)), first split it into partial fractions.
Expand each fraction as a binomial with power $- 1$ −1, then add like terms.
Each piece has its own interval of validity; the whole expansion is valid only where both are, so take the narrower interval.

Calculate

?

f (x) = \frac{4 - 5 x}{( 1 + x ) ( 2 - x )}

. Given that

f (x) = \frac{3}{1 + x} - \frac{2}{2 - x}

, find the coefficient of

x^{2}

in the expansion of

f (x)

in ascending powers of

x

. Give your answer as a decimal.

Sort it

?Is the expansion valid for the value of

x

given?

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

Mission check

?

(1 + x)^{- 1} = 1 - x + x^{2} - \dots

Why is it nonsense for

x = 2

?

Mission 3

Add every whole number from 1 to 100 in seconds. How?

Make a guess: Add every whole number from 1 to 100 in seconds. How?

Sequences and series · arithmetic series

Arithmetic sequences (A2 only)

An arithmetic sequence (or arithmetic progression, AP) goes up or down by the same common difference $d$ d each time.
With first term $a$ a, the $n$ nth term is

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

Example: $7, 11, 15, \dots$ 7, 11, 15, … has $a = 7$ a=7, $d = 4$ d=4, so $u_{n} = 7 + 4 (n - 1) = 4 n + 3$ uₙ=7+4(n−1)=4n+3.
The $n$ nth term has only $n - 1$ n−1 steps of $d$ d added to $a$ a.

Calculate

?Find the 20th term of the arithmetic sequence

5, 8, 11, \dots

Sort it

?Is the sequence arithmetic?

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

Sequences and series · arithmetic series

The sum to $n$ n terms, and its proof

Write the series forwards and backwards, then add:

$S_{n} = a + (a + d) + \dots + (l - d) + l$ Sₙ=a+(a+d)+…+(l−d)+l $S_{n} = l + (l - d) + \dots + (a + d) + a$ Sₙ=l+(l−d)+…+(a+d)+a

Every column adds to $a + l$ a+l, and there are $n$ n columns: $2 S_{n} = n (a + l)$ 2Sₙ=n(a+l).
So $S_{n} = \frac{n}{2} (a + l)$ Sₙ=n/2(a+l). With $l = a + (n - 1) d$ l=a+(n−1)d: $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ Sₙ=n/2(2a+(n−1)d).
With $a = 1$ a=1, $d = 1$ d=1: $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$ 1+2+…+n=(n(n+1))/2.

Worked example Sequences and series · arithmetic series

Finding $a$ a and $d$ d first

Question

The 4th term of an arithmetic series is 17 and the 11th term is 45. Find the sum of the first 20 terms.

1
Write the two terms with $u_{n} = a + (n - 1) d$ uₙ=a+(n−1)d: $a + 3 d = 17$ a+3d=17 and $a + 10 d = 45$ a+10d=45
2
Subtract to find $d$ d: $7 d = 28$ 7d=28, so $d = 4$ d=4, then $a = 17 - 12 = 5$ a=17−12=5
3
Use the sum formula: $S_{20} = \frac{20}{2} (2 (5) + 19 (4)) = 10 (10 + 76)$ S₂₀=20/2(2(5)+19(4))=10(10+76)

Answer

S_{20} = 860

⚠️ Watch out

the 4th term is

a + 3 d

, not

a + 4 d

; an off-by-one here spoils both

a

and the sum.

Sequences and series · arithmetic series

Sigma notation

$\sum_{r = 1 n} f (r)$ Σ_(r=1)ⁿ f(r) means "add $f (r)$ f(r) for $r = 1, 2, \dots, n$ r=1, 2, …, n".
If $f (r)$ f(r) is linear, such as $3 r + 2$ 3r+2, the terms form an AP: write out the first and last terms, then use $S_{n} = \frac{n}{2} (a + l)$ Sₙ=n/2(a+l).
The number of terms is (top − bottom + 1): $\sum_{r = 5 20}$ Σ_(r=5)²⁰ has 16 terms.
$\sum_{r = 1 n} 1 = n$ Σ_(r=1)ⁿ 1=n, because you add 1 exactly $n$ n times.

Quick check

?Evaluate

r = 1 \sum 20 (3 r + 2)

.

Calculate

?Asha saves £50 in the first month of a savings plan, and each month she saves £5 more than the month before. In which month does her total saving first exceed £3,000?

Mission check

?How can you add every whole number from 1 to 100 in seconds?

Mission 4

Halve the distance to a wall again and again. Do you ever arrive?

Make a guess: Halve the distance to a wall again and again. Do you ever arrive?

Sequences and series · geometric series

Geometric sequences (A2 only)

A geometric sequence multiplies by the same common ratio $r$ r each time: $r = \frac{u _{n + 1}}{u _{n}}$ r = (uₙ₊₁)/uₙ.
With first term $a$ a the terms are $a, a r, a r^{2}, a r^{3}, \dots$ a, ar, ar², ar³, … so

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

$3, 6, 12, 24, \dots$ 3, 6, 12, 24, … has $r = 2$ r = 2. $40, - 20, 10, - 5, \dots$ 40, −20, 10, −5, … has $r = - \frac{1}{2}$ r = −1/2: a negative ratio makes the signs alternate.
It is also called a geometric progression (GP).

Worked example Sequences and series · geometric series

Finding a term from two other terms

Question

The 3rd term of a geometric sequence is 12 and the 6th term is 96. Find the 10th term.

1
Write both terms with $u_{n} = a r^{n - 1}$ uₙ = arⁿ⁻¹: $a r^{2} = 12$ ar² = 12 and $a r^{5} = 96$ ar⁵ = 96
2
Divide the equations to eliminate $a$ a: $r^{3} = 8$ r³ = 8, so $r = 2$ r = 2
3
Substitute back: $4 a = 12$ 4a = 12, so $a = 3$ a = 3
4
Use the formula with $n = 10$ n = 10: $u_{10} = 3 \times 2^{9}$ u₁₀ = 3 × 2⁹

Answer

u_{10} = 1536

⚠️ Watch out

the power is

n - 1

. Working out

3 \times 2^{10} = 3072

gives the 11th term.

Sequences and series · geometric series

The sum of a finite geometric series (A2 only)

Write $S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}$ Sₙ = a + ar + ar² + … + arⁿ⁻¹.
Multiply by $r$ r: $r S_{n} = a r + a r^{2} + \dots + a r^{n - 1} + a r^{n}$ rSₙ = ar + ar² + … + arⁿ⁻¹ + arⁿ.
Subtract: every middle term cancels, leaving $S_{n} - r S_{n} = a - a r^{n}$ Sₙ − rSₙ = a − arⁿ. Factorise and divide by $1 - r$ 1 − r:

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}, r \neq = 1

The same formula as $\frac{a ( r ^{n} - 1 )}{r - 1}$ (a(rⁿ − 1))/(r − 1) is tidier when $r > 1$ r > 1.
Learn this proof: you can be asked to prove the formula.

Calculate

?Find the sum of the first 10 terms of the series

5 + 15 + 45 + \dots

Give the exact value.

Worked example Sequences and series · geometric series

Using logarithms to find n

Question

A runner covers 20 km in week 1 of a training plan and increases her weekly distance by 5% each week. In which week does her total distance first exceed 1,000 km?

1
The weekly distances are geometric with $a = 20$ a = 20, $r = 1.05$ r = 1.05: $S_{n} = \frac{20 ( 1.0 5 ^{n} - 1 )}{0.05} = 400 (1.0 5^{n} - 1)$ Sₙ = (20(1.05ⁿ − 1))/0.05 = 400(1.05ⁿ − 1)
2
Set up the inequality: $400 (1.0 5^{n} - 1) > 1000$ 400(1.05ⁿ − 1) > 1000, so $1.0 5^{n} > 3.5$ 1.05ⁿ > 3.5
3
Take logs of both sides: $n ln 1.05 > ln 3.5$ n ln 1.05 > ln 3.5, so $n > \frac{ln 3.5}{ln 1.05} = 25.67 \dots$ n > (ln 3.5)/(ln 1.05) = 25.67…
4
$n$ n is a whole number of weeks, so round up and check: $S_{25} = 954.5$ S₂₅ = 954.5 km and $S_{26} = 1022.3$ S₂₆ = 1022.3 km

Answer

week 26

⚠️ Watch out

always round up here.

n = 25

still gives a total below 1,000 km.

Sequences and series · geometric series

Convergence and the sum to infinity (A2 only)

If $∣ r ∣ < 1$ |r| < 1, then $r^{n} \to 0$ rⁿ → 0 as $n \to \infty$ n → ∞, so $S_{n}$ Sₙ gets closer and closer to $\frac{a}{1 - r}$ a/(1 − r). The series converges, and its sum to infinity is this limit of $S_{n}$ Sₙ (no finite $S_{n}$ Sₙ need equal it).

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

$∣ r ∣ < 1$ |r| < 1 is modulus notation for $- 1 < r < 1$ −1 < r < 1.
If $∣ r ∣ \geq 1$ |r| ≥ 1 the series diverges: it has no sum to infinity.
In the diagram, $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ 1/2 + 1/4 + 1/8 + … fills the whole unit square: $S_{\infty} = \frac{1/2}{1 - 1/2} = 1$ S_(∞) = (1/2)/(1 − 1/2) = 1.

Sort it

?Does each geometric series have a sum to infinity?

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

Quick check

?Find the sum to infinity of

24 - 12 + 6 - 3 + \dots

Calculate

?The second term of a geometric series is 6. The sum to infinity of the series is 4 times its first term. Find the sum to infinity.

Mission check

?You start 4 m from a wall and step 2 m, then 1 m, then 0.5 m, always half the gap left. Do you ever arrive?

Mission 5

$u_{n + 1} = \frac{1}{1 - u _{n}}$ uₙ₊₁ = 1/(1−uₙ) and $u_{1} = 2$ u₁ = 2. What is $u_{100}$ u₁₀₀?

Make a guess: $u_{n + 1} = \frac{1}{1 - u _{n}}$ uₙ₊₁ = 1/(1−uₙ) and $u_{1} = 2$ u₁ = 2. What is $u_{100}$ u₁₀₀?

Sequences and series · recurrence and sigma

Two ways to define a sequence (A2 only)

Formula for the nth term: $u_{n} = 3 n + 1$ uₙ = 3n + 1 gives 4, 7, 10, … and jumps straight to any term.
Recurrence relation: a rule $u_{n + 1} = f (u_{n})$ uₙ₊₁ = f(uₙ) plus a first term. $u_{n + 1} = 2 u_{n} - 1$ uₙ₊₁ = 2uₙ − 1 with $u_{1} = 3$ u₁ = 3 gives 3, 5, 9, 17, …
The first term matters: the same rule with $u_{1} = 1$ u₁ = 1 gives 1, 1, 1, …
A sequence is a list of terms; a series is a sum of terms. A sequence can be finite or infinite.

Worked example Sequences and series · recurrence and sigma

Finding a constant in a recurrence relation

Question

A sequence is defined by

u_{n + 1} = k u_{n} + 3

,

u_{1} = 2

, where

k

is a positive constant. Given that

u_{3} = 17

, find

k

and

u_{4}

.

1
Generate terms in terms of $k$ k: $u_{2} = 2 k + 3$ u₂ = 2k + 3
2
Apply the rule again: $u_{3} = k (2 k + 3) + 3 = 2 k^{2} + 3 k + 3$ u₃ = k(2k + 3) + 3 = 2k² + 3k + 3
3
Set equal to 17 and solve: $2 k^{2} + 3 k - 14 = 0$ 2k² + 3k − 14 = 0, so $(k - 2) (2 k + 7) = 0$ (k − 2)(2k + 7) = 0
4
$k > 0$ k > 0, so reject $k = - \frac{7}{2}$ k = −7/2: $k = 2$ k = 2. Then $u_{4} = 2 \times 17 + 3$ u₄ = 2 × 17 + 3

Answer

k = 2

and

u_{4} = 37

⚠️ Watch out

u_{3}

comes from

u_{2}

, not from

u_{1}

. Writing

u_{3} = 2 k + 3

skips a step.

Sequences and series · recurrence and sigma

Increasing, decreasing and periodic sequences (A2 only)

Increasing: $u_{n + 1} > u_{n}$ uₙ₊₁ > uₙ for all $n$ n. Decreasing: $u_{n + 1} < u_{n}$ uₙ₊₁ < uₙ for all $n$ n.
Periodic: the terms repeat in a cycle. If $u_{n + k} = u_{n}$ uₙ₊ₖ = uₙ for all $n$ n, with $k$ k the smallest such number, the sequence has order $k$ k.
$u_{n + 1} = \frac{1}{u _{n}}$ uₙ₊₁ = 1/uₙ, $u_{1} = 3$ u₁ = 3 gives $3, \frac{1}{3}, 3, \frac{1}{3}, \dots$ 3, 1/3, 3, 1/3, …: periodic, order 2.
Long-term behaviour: a sequence may converge to a limit, diverge, or oscillate.

Sort it

?Classify each sequence.

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

Sequences and series · recurrence and sigma

Sigma notation (A2 only)

$r = 1 \sum n f (r)$ Σ_(r=1)ⁿ f(r) means: substitute $r = 1, 2, \dots, n$ r = 1, 2, …, n and add the results.
$r = 1 \sum 4 (2 r + 1) = 3 + 5 + 7 + 9 = 24$ Σ_(r=1)⁴ (2r + 1) = 3 + 5 + 7 + 9 = 24
Count the terms carefully: $r = 3 \sum 10$ Σ_(r=3)¹⁰ has $10 - 3 + 1 = 8$ 10 − 3 + 1 = 8 terms.
Adding 1 for each of $n$ n terms: $r = 1 \sum n 1 = n$ Σ_(r=1)ⁿ 1 = n.

Calculate

?Find the value of

r = 5 \sum 20 (4 r - 3)

.

Worked example Sequences and series · recurrence and sigma

Modelling with a recurrence relation

Question

A lake is stocked with 1,000 fish. Each year 10% of the fish die and then 200 new fish are added. Model this with a recurrence relation, find the number of fish at the start of year 4, and the long-term number.

1
Keep 90% and add 200: $P_{n + 1} = 0.9 P_{n} + 200$ Pₙ₊₁ = 0.9Pₙ + 200, $P_{1} = 1000$ P₁ = 1000
2
Generate terms: $P_{2} = 1100$ P₂ = 1100, $P_{3} = 1190$ P₃ = 1190, $P_{4} = 1271$ P₄ = 1271
3
If the sequence converges to $L$ L, then $P_{n}$ Pₙ and $P_{n + 1}$ Pₙ₊₁ both tend to $L$ L, so $L = 0.9 L + 200$ L = 0.9L + 200
4
Solve: $0.1 L = 200$ 0.1L = 200, so $L = 2000$ L = 2000

Answer

1,271 fish at the start of year 4; the population approaches 2,000

⚠️ Watch out

the model assumes the death rate and restocking never change. Real numbers of fish are whole, so treat the model's values as estimates.

Quick check

?A population is modelled by

P_{n + 1} = 0.8 P_{n} + 300

. The sequence converges. What value does it approach?

Calculate

?A sequence is defined by

u_{1} = 2

and

u_{n + 1} = 5 - u_{n}

. Find

r = 1 \sum 25 u_{r}

.

Mission check

?

u_{n + 1} = \frac{1}{1 - u _{n}}

and

u_{1} = 2

. What is

u_{100}

?

Recap

Big ideas

Row $n$ n of Pascal's triangle, or $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ ⁿCᵣ=(n!)/(r! (n−r)!), gives the coefficients of $(a + b)^{n}$ (a+b)ⁿ.
The general term is $(r n) a^{n - r} b^{r}$ C(n, r)a^(n−r)b^r: raise the whole of $b$ b, sign and number included, to the power $r$ r.
With a small $x$ x, the first few terms of an expansion give an accurate estimate.
For rational $n$ n, $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$ (1+x)ⁿ=1+nx+(n(n−1))/(2!)x²+… is infinite and valid only for $∣ x ∣ < 1$ |x|<1.
For $(a + b x)^{n}$ (a+bx)ⁿ take out $a^{n}$ aⁿ first; the expansion is then valid for $\frac{b x}{a} < 1$ |bx/a|<1.
With partial fractions, expand each piece and keep the narrower interval of validity.
An arithmetic sequence has a common difference $d$ d; its $n$ nth term is $u_{n} = a + (n - 1) d$ uₙ=a+(n−1)d.
Reverse-and-add proves $S_{n} = \frac{n}{2} (a + l) = \frac{n}{2} (2 a + (n - 1) d)$ Sₙ=n/2(a+l)=n/2(2a+(n−1)d); in particular $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$ 1+2+…+n=(n(n+1))/2.
For $\sum$ Σ with a linear expression, find the first term, last term and number of terms, then use the sum formula.
Geometric sequence: $u_{n} = a r^{n - 1}$ uₙ = arⁿ⁻¹. Finite sum: $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ Sₙ = (a(1 − rⁿ))/(1 − r), proved by subtracting $r S_{n}$ rSₙ from $S_{n}$ Sₙ.
If $∣ r ∣ < 1$ |r| < 1 the series converges and $S_{\infty} = \frac{a}{1 - r}$ S_(∞) = a/(1 − r); if $∣ r ∣ \geq 1$ |r| ≥ 1 it diverges.
To find $n$ n from a sum, form an inequality, take logs and round up.
A sequence can be defined by an nth-term formula or by a recurrence relation $u_{n + 1} = f (u_{n})$ uₙ₊₁ = f(uₙ) with a first term.
Sequences can be increasing, decreasing or periodic (order = length of the cycle); a converging recurrence has limit $L = f (L)$ L = f(L).
$\sum_{r = a b}$ Σ_(r=a)^b has $b - a + 1$ b − a + 1 terms; $\sum_{r = 1 n} 1 = n$ Σ_(r=1)ⁿ 1 = n.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Sequences and series for CCEA GCE Mathematics (2210). 🎉

Mission 1How can you work out $1.0 1^{10}$ 1.01¹⁰ to 4 decimal places in your head?
Mission 2 $(1 + x)^{- 1} = 1 - x + x^{2} - \dots$ (1+x)⁻¹=1−x+x²−… Why is it nonsense for $x = 2$ x=2?
Mission 3Add every whole number from 1 to 100 in seconds. How?
Mission 4Halve the distance to a wall again and again. Do you ever arrive?
Mission 5 $u_{n + 1} = \frac{1}{1 - u _{n}}$ uₙ₊₁ = 1/(1−uₙ) and $u_{1} = 2$ u₁ = 2. What is $u_{100}$ u₁₀₀?

Your stars: 0 / 0

Next: test yourself in the Verify stage.

📣 Smashed it? Share your score

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

→ Back to A-Level Mathematics · CCEA

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

Sequences and series

How can you work out 1.01101.01¹⁰ to 4 decimal places in your head?

Pascal's triangle and (a+b)n(a+b)ⁿ

Factorials and nCr​ⁿCᵣ

Expanding a bracket in full

Using an expansion to approximate

Estimating a power

(1+x)−1=1−x+x2−…(1+x)⁻¹=1−x+x²−… Why is it nonsense for x=2x=2?

The expansion for any rational nn (A2 only)

A negative power

Expanding (a+bx)n(a+bx)ⁿ

Taking out a factor, then approximating

Combining with partial fractions

Add every whole number from 1 to 100 in seconds. How?

Arithmetic sequences (A2 only)

The sum to nn terms, and its proof

Finding aa and dd first

Sigma notation

Halve the distance to a wall again and again. Do you ever arrive?

Geometric sequences (A2 only)

Finding a term from two other terms

The sum of a finite geometric series (A2 only)

Using logarithms to find n

Convergence and the sum to infinity (A2 only)

un+1​=1−un​1​uₙ₊₁ = 1/(1−uₙ) and u1​=2u₁ = 2. What is u100​u₁₀₀?

Two ways to define a sequence (A2 only)

Finding a constant in a recurrence relation

Increasing, decreasing and periodic sequences (A2 only)

Sigma notation (A2 only)

Modelling with a recurrence relation

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

How can you work out $1.0 1^{10}$ 1.01¹⁰ to 4 decimal places in your head?

Pascal's triangle and $(a + b)^{n}$ (a+b)ⁿ

Factorials and $^{n} C_{r}$ ⁿCᵣ

$(1 + x)^{- 1} = 1 - x + x^{2} - \dots$ (1+x)⁻¹=1−x+x²−… Why is it nonsense for $x = 2$ x=2?

The expansion for any rational $n$ n (A2 only)

Expanding $(a + b x)^{n}$ (a+bx)ⁿ

The sum to $n$ n terms, and its proof

Finding $a$ a and $d$ d first

$u_{n + 1} = \frac{1}{1 - u _{n}}$ uₙ₊₁ = 1/(1−uₙ) and $u_{1} = 2$ u₁ = 2. What is $u_{100}$ u₁₀₀?