Sequences & Series

9 methods · 24 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

The full chapter (9 methods · 24 worked questions) comes with Platform Access. See pricing

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 (3)

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 $39$

Worked solution

Answer:B

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

38

, each summing to

0

. The remaining term is

a_{39} = - 3

. Therefore

n = 1 \sum 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 $x^{10} + 10 x^{9} + (10 \times 9) x^{8} + \dots + (10 \times 9 \times \dots \times 2) x + (10 \times 9 \times \dots \times 2 \times 1)$

Worked solution

Answer:C

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 $7$

Worked solution

Answer:A

Compute

x_{4}

x_{4} = \frac{23 \times 1 - 53}{5 \times 1 + 1} = \frac{- 30}{6} = - 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.