Sequences & Series | The Maths Tailor — UK Methods

Sequences & Series

8 methods · 21 questions

The instant you meet a rule of the form $u_{n + 1} = f (u_{n})$ (or a term built from $(- 1)^{n}$ , a derivative, a square root, etc.), write out $u_{1}, u_{2}, u_{3}, u_{4}, u_{5}$ explicitly. The pattern you need (periodicity, a closed form, a constant ratio) almost always becomes visible in the first handful of terms, and this costs seconds.

Trigger: any sequence given by a recurrence or an index-dependent formula, especially when the question asks for a far-off term or a long sum.

Instances:
(i) an alternating $(- 1)^{n}$ rule reveals a short repeating block;
(ii) a rational map $u_{n + 1} = \frac{a u _{n} + b}{c u _{n} + d}$ reveals a cycle;
(iii) a square-root or derivative recurrence reveals exponents in arithmetic progression.

Linked questions (6)

TMUA 2016 Paper 1 — Q4

The sequence

a_{n}

is defined by the rule:

a_{n} = (- 1)^{n} - (- 1)^{n - 1} + (- 1)^{n + 2} for n \geq 1.

Find the value of

n = 1 \sum 39 a_{n}

.

(A)

- 39

(B)

- 3

(C)

- 1

(D)

0

(E)

1

(F)

3

(G)

39

Voir la correction

Compute the first few terms. For

n = 1

:

a_{1} = (- 1) - 1 + (- 1) = - 3

. For

n = 2

:

a_{2} = 1 - (- 1) + 1 = 3

. For

n = 3

:

a_{3} = (- 1) - 1 + (- 1) = - 3

. For

n = 4

:

a_{4} = 1 - (- 1) + 1 = 3

. The sequence alternates

- 3, 3, - 3, 3, \dots

Each consecutive pair sums to zero:

a_{2 k - 1} + a_{2 k} = 0

. There are 19 such pairs covering

n = 1

to

38

, each summing to

0

. The remaining term is

a_{39} = - 3

. Therefore

\sum_{n = 1}^{39} a_{n} = 0 + (- 3) = - 3

. Hence the answer is B.

TMUA 2016 Paper 2 — Q6

The sequence of functions

f_{1} (x), f_{2} (x), f_{3} (x), \dots

is defined as follows:

f_{1} (x) = x^{10}

f_{n + 1} (x) = x f_{n}^{'} (x) for n \geq 1

where

f_{n}^{'} (x) = \frac{d f _{n} ( x )}{d x}

.

Find the value of

n = 1 \sum 20 f_{n} (x)

(A)

\frac{x ^{10} ( x ^{20} - 1 )}{x - 1}

(B)

\frac{x ^{10} ( x ^{21} - 1 )}{x - 1}

(C)

(\frac{1 0 ^{20} - 1}{9}) x^{10}

(D)

(\frac{1 0 ^{21} - 1}{9}) x^{10}

(E)

(\frac{( 10 x ) ^{20} - 1}{10 x - 1}) x^{10}

(F)

(\frac{( 10 x ) ^{21} - 1}{10 x - 1}) x^{10}

(G)

x^{10} + x^{9} + x^{8} + \dots + x + 1

(H)

x^{10} + 10 x^{9} + (10 \times 9) x^{8} + \dots + (10 \times 9 \times \dots \times 2) x + (10 \times 9 \times \dots \times 2 \times 1)

Voir la correction

Compute the first few terms of the sequence.

f_{1} (x) = x^{10}

.

f_{2} (x) = x \cdot f_{1}^{'} (x) = x \cdot 10 x^{9} = 10 x^{10}

.

f_{3} (x) = x \cdot f_{2}^{'} (x) = x \cdot 100 x^{9} = 100 x^{10} = 1 0^{2} x^{10}

. In general,

f_{n} (x) = 1 0^{n - 1} x^{10}

. Therefore:

n = 1 \sum 20 f_{n} (x) = (1 + 10 + 1 0^{2} + \dots + 1 0^{19}) x^{10} = \frac{1 0 ^{20} - 1}{10 - 1} x^{10} = (\frac{1 0 ^{20} - 1}{9}) x^{10} .

Hence the answer is C.

TMUA 2017 Paper 1 — Q11

The sequence

x_{n}

is defined by the rules

x_{1} = 7, x_{n + 1} = \frac{23 x _{n} - 53}{5 x _{n} + 1} .

The first three terms in the sequence are 7, 3, 1. What is the value of

x_{100}

?

(A)

- 5

(B)

0

(C)

1

(D)

3

(E)

7

Voir la correction

Compute

x_{4}

:

- 5

Compute

x_{5}

:

x_{5} = \frac{23 \times ( - 5 ) - 53}{5 \times ( - 5 ) + 1} = \frac{- 168}{- 24} = 7.

So the sequence cycles with period 4:

7, 3, 1, - 5, 7, 3, 1, - 5, \dots

Since

100 = 25 \times 4

, we have

x_{100} = x_{4} = - 5

. Hence the answer is A.

TMUA 2019 Paper 1 — Q4

The sequence

x_{n}

is given by:

x_{1} = 10

x_{n + 1} = x_{n} for n \geq 1

What is the value of

x_{100}

? [Note that

a^{b^{c}}

means

a^{(b^{c})}

]

(A)

1 0^{2^{99}}

(B)

1 0^{2^{100}}

(C)

1 0^{2^{- 99}}

(D)

1 0^{2^{- 100}}

(E)

1 0^{- 2^{99}}

(F)

1 0^{- 2^{100}}

(G)

1 0^{- 2^{- 99}}

(H)

1 0^{- 2^{- 100}}

Voir la correction

We have

x_{1} = 10 = 1 0^{1}

. Then

x_{2} = 10 = 1 0^{1/2} = 1 0^{2^{- 1}}

. Next,

x_{3} = x_{2} = 1 0^{2^{- 1} \cdot 1/2} = 1 0^{2^{- 2}}

. In general,

x_{n} = 1 0^{2^{- (n - 1)}}

. Therefore

x_{100} = 1 0^{2^{- 99}}

. Hence the answer is C.

TMUA 2022 Paper 1 — Q5

The terms

x_{n}

of a sequence follow the rule

x_{n + 1} = \frac{x _{n} + p}{x _{n} + q}

where

p

and

q

are real numbers. Given that

x_{1} = 3

,

x_{2} = 5

, and

x_{3} = 7

, find the value of

x_{4}

.

(A)

- 5

(B)

5

(C)

\frac{51}{7}

(D)

\frac{15}{2}

(E)

\frac{23}{3}

(F)

9

(G)

11

(H)

13

Voir la correction

Using

x_{2} = \frac{x _{1} + p}{x _{1} + q}

:

5 = \frac{3 + p}{3 + q}

, so

15 + 5 q = 3 + p

, giving

p - 5 q = 12

. Using

x_{3} = \frac{x _{2} + p}{x _{2} + q}

:

7 = \frac{5 + p}{5 + q}

, so

35 + 7 q = 5 + p

, giving

p - 7 q = 30

. Subtracting the first equation from the second:

- 2 q = 18

, so

35 + 7 q = 5 + p

. Then

p = 12 + 5 (- 9) = 12 - 45 = - 33

. Now

x_{4} = \frac{7 + p}{7 + q} = \frac{7 - 33}{7 - 9} = \frac{- 26}{- 2} = 13

. Hence the answer is H.

TMUA Specimen Paper 2 — Q16

The sequence

a_{n}

is given by the rule:

a_{1} = 2

and

a_{n + 1} = a_{n} + (- 1)^{n}

for

n \geq 1

. What is

n = 1 \sum 100 a_{n}

?

(A) 150
(B) 250
(C)

- 4750

(D) 5150
(E)

4 (1 - (\frac{1}{2})^{100})

(F)

- 4750

Voir la correction

Compute the first few terms:

a_{1} = 2

,

a_{2} = 2 + (- 1)^{1} = 1

,

a_{3} = 1 + (- 1)^{2} = 2

,

a_{4} = 2 + (- 1)^{3} = 1

,

a_{5} = 1 + (- 1)^{4} = 2

. The sequence alternates

2, 1, 2, 1, \dots

with period 2. Each pair

(a_{2 k - 1}, a_{2 k}) = (2, 1)

sums to 3. There are 50 such pairs in the first 100 terms, so

n = 1 \sum 100 a_{n} = 50 \times 3 = 150

. Hence the answer is A.