Learning Goal
Syllabus Requirement
Students need to learn about
sequences, with the below point being highlighted in particular:

Sequences given by a formula for the ??n^{th}?? term; and

Sequences generated by a simple relation of the form ??x_{n+1}=f(x_{n})??.
Study Notes
A sequences is a series of numbers following a set rule, such as ??2, \: 4, \:, 6, \:, 8\: ...??. Each number or expression
in a sequence is called a term.
The notation ??U_{n}?? is often used to refer to the ??n^{th}?? term of a sequence. So ??U_{1}?? is the first term, ??U_{2}?? the second term and so on.
There are two ways in which a general sequence is usually specified:
The 2 ways of specifying a general sequence.
1. Formula for the ??n^{th}?? term.
If a formula for the ??n^{th}?? term is given, then any term in the sequence can be found by substituting the position of the
term you require into the formula.
2. Using a simple relationship between terms (also known as a 'recurrence relation').
A recurrence relation can define a sequence, often by giving the first term of the sequence and a relationship that links any term
to the previous term.
Worked Examples
We'll now run through some examples of each of the two ways of specifying a general sequence.
1. Formula for the ??n^{th}?? term.
Example 1
Question
The ??n^{th}?? term of a sequence is given by ??u_{n} = 3n + 1??. What is the ??10^{th}?? term?
Show Solution
Worked Solution
Substitute ??n=10?? into the formula for the ??n^{th}?? term:
$$U_{n} = 3n + 1$$
$$\implies U_{10} = 3 \times 10 +1$$
$$\implies U_{10} = 30 + 1$$
$$\implies U_{10} = 31$$
Example 2
Question
The ??n^{th}?? term of a sequence is given by ??u_{n} = \frac{n^{2}}{n+1}??. What is the ??3^{rd}?? term?
Show Solution
Worked Solution
Substitute ??n=3?? into the formula for the ??n^{th}?? term:
$$U_{n} = \frac{n^{2}}{n+1}$$
$$\implies U_{3} = \frac{3^{2}}{3 + 1}$$
$$\implies U_{3} = \frac{9}{4}$$
Example 3
Question
Find the value of ??n??, where the ??n^{th}?? term is given by ??2n  9??, where ??U_{n} = 13??.
Show Solution
Worked Solution
Substitute ??U_{n} = 13?? into the formula for the ??n^{th}?? term:
$$U_{n} = 2n9$$
$$\implies 13 = 2n9$$
$$\implies 13+9 = 2n$$
$$\implies 22 = 2n$$
$$\implies n = 11$$
The ??11^{th}?? term in the sequence is ??13??.
2. Using a simple relationship between terms (also known as a 'recurrence relation').
Example 4
Question
If ??U_{1} = 3?? and the recurrence relation is ??U_{n+1} = 3U_{n}  7??, find the first ??3?? terms of the sequence.
Show Solution
Worked Solution
We have ??U_{1}?? so let's start with ??U_{2}??:
$$U_{n+1} = 3U_{n}  7$$
$$\implies U_{2} = 3U_{1}  7$$
$$\implies U_{2} = 3 \times 3  7$$
$$\implies U_{2} = 2$$
We now know that ??U_{2} = 2?? and therefore we can find ??U_{3}??:
$$U_{n+1} = 3U_{n}  7$$
$$\implies U_{3} = 3U_{2}  7$$
$$\implies U_{3} = 3 \times 2  7$$
$$\implies U_{3} = 1 $$
Therefore the first ??3?? terms of the sequence are ??3, \: 2, \: 1??.
Example 5
Question
If ??U_{1} = 5?? and ??U_{2} = 8??, find ??m?? in the recurrence relation ??U_{n+1} = mU_{n}  2??.
Show Solution
Worked Solution
We can write an equation based on the recurrence relation and solve for ??m??:
$$U_{n+1} = mU_{n}  2$$
$$\implies U_{2} = mU_{1}  2$$
$$\implies 8 = 5m  2$$
$$\implies 8 + 2 = 5m$$
$$\implies 10 = 5m$$
$$\implies m = 2$$
Therefore the recurrence relation is ??U_{n+1} = 2U_{n}  2??.
Exam Questions
The table below contains every exam question that has been asked on this topic, this includes normal papers from both January
and June sittings, International papers and Specimen papers.
To see an exam question and solution simply click on the load question icon (), the question will appear below the table and
the solution can be shown by clicking the "Show Solution" button that also appears.
The full exam paper and mark schemes are also available for download by clicking on the download icons in each
row of the table ().
Exam Board 
Subject 
Paper 
Year 
Month 
Module 
Question No. 
Parts 
Total Marks 
Load Question 
Exam Paper 
Mark Scheme 
Edexcel 
Maths 
Standard 
2006 
January 
Core 1 
2 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2006 
June 
Core 1 
4 
(a), (b) 
5 



Edexcel 
Maths 
Standard 
2007 
June 
Core 1 
8 
(a), (b), (c) 
7 



Edexcel 
Maths 
Standard 
2008 
January 
Core 1 
7 
(a), (b), (c), (d) 
8 



Edexcel 
Maths 
Standard 
2008 
June 
Core 1 
5 
(a), (b), (c) 
6 



Edexcel 
Maths 
Standard 
2009 
June 
Core 1 
7 
(a), (b), (c) 
7 



Edexcel 
Maths 
Standard 
2010 
June 
Core 1 
5 
(a), (b) 
4 



Edexcel 
Maths 
Standard 
2011 
January 
Core 1 
4 
(a), (b) 
5 



Edexcel 
Maths 
Standard 
2011 
June 
Core 1 
5 
(a), (b), (c) 
7 



Edexcel 
Maths 
Standard 
2012 
January 
Core 1 
4 
(a), (b), (c) 
6 



Edexcel 
Maths 
Standard 
2012 
June 
Core 1 
5 
(a), (b), (c) 
7 



Edexcel 
Maths 
Standard 
2013 
January 
Core 1 
4 
(a), (b) 
5 



Edexcel 
Maths 
Standard 
2013 
June 
Core 1 
4 
(a), (b) 
7 



Edexcel 
Maths 
International 
2013 
June 
Core 1 
6 
(a), (b), (c), (d) 
9 



Edexcel  Maths  Standard  2006  January  Core 1  Question 2
Question
2.
The sequence of positive numbers ??u_{1}, \: u_{2}, \:, u_{3}, \: ...??, is given by:
$$u_{n+1} = (u_{n}  3)^{2}, \: \: \: u_{1}=1.$$
2. (a)
Find ??u_{2}??, ??u_{3}?? and ??u_{4}??. [3]
2. (b)
Write down the value of ??u_{20}??. [1]
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Show Solution
Worked Solution
2. (a)
We have a recurrence relation here, so can substitute values in to get each term, in turn. Starting with ??u_{2}??:
$$u_{n+1} = (u_{n}3)^{2}$$
$$\implies u_{2} = (u_{1}3)^{2}$$
$$\implies u_{2} = (13)^{2}$$
$$\implies u_{2} = (2)^{2}$$
$$\implies u_{2} = 4$$
Now we can find ??u_{3}??:
$$u_{n+1} = (u_{n}3)^{2}$$
$$\implies u_{3} = (u_{2}3)^{2}$$
$$\implies u_{3} = (43)^{2}$$
$$\implies u_{3} = 1^{2}$$
$$\implies u_{3} = 1$$
And finally we can find ??u_{4}??:
$$u_{n+1} = (u_{n}3)^{2}$$
$$\implies u_{4} = (u_{3}3)^{2}$$
$$\implies u_{4} = (13)^{2}$$
$$\implies u_{4} = (2)^{2}$$
$$\implies u_{4} = 4$$
2. (b)
Start by noting that the question asks you to write down the answer, which means it should not take very much, if any, working at all.
We can see, from 2.(b) that each subsequent terms will alternate between ??1?? and ??4??, because the recurrence relation always
stays the same and because we can see the cyclic pattern in our answer to 2. (a). Noting that all of the evennumbered terms are ??4??
means that ??u_{20}?? will also be ??4?? (as it is even).
Edexcel  Maths  Standard  2006  June  Core 1  Question 4
Question
4.
A sequence ??a_{1}, \: a_{2}, \: a_{3}, \: ...?? is defined by:
$$a_{1} = 3,$$
$$a_{n+1} = 3a_{n} 5, \: \: \: n \geq 1.$$
4. (a)
Find the value of ??a_{2}?? and the value of ??a_{3}??. [2]
4. (b)
Calculate the value of ??\sum\limits_{r=1}^5 a_{r}??. [3]
Hide Question
Show Solution
Worked Solution
4. (a)
We have a recurrence relation here, so can substitute values in to get each term, in turn. Starting with ??a_{2}??:
$$a_{n+1} = 3a_{n}5$$
$$\implies a_{2} = 3a_{1} 5$$
$$\implies a_{2} = 3 \times 3 5$$
$$\implies a_{2} = 9  5$$
$$\implies a_{2} = 4$$
Now we can find ??a_{3}??:
$$a_{n+1} = 3a_{n}5$$
$$\implies a_{3} = 3a_{2}  5$$
$$\implies a_{3} = 3 \times 4  5$$
$$\implies a_{3} = 12  5$$
$$\implies a_{3} = 7$$
4. (b)
We're being asked to calculate the sum of the terms from ??a_{1}?? to ??a_{5}?? The easiest way of doing this is to calculate ??a_{4}?? and ??a_{5}??. Let's start
by finding ??a_{4}??:
$$a_{n+1} = 3a_{n}5$$
$$\implies a_{4} = 3a_{3}  5$$
$$\implies a_{4} = 3 \times 7  5$$
$$\implies a_{4} = 21  5$$
$$\implies a_{4} = 16$$
Now we can find ??a_{5}??:
$$a_{n+1} = 3a_{n}5$$
$$\implies a_{5} = 3a_{4}  5$$
$$\implies a_{5} = 3 \times 16  5$$
$$\implies a_{5} = 48  5$$
$$\implies a_{5} = 43$$
Finally, we can calculate the required sum:
$$\sum\limits_{r=1}^5 a_{r}$$
$$= a_{1} + a_{2} + a_{3} + a_{4} + a_{5}$$
$$= 3 + 4 + 7 + 16 + 43$$
$$= 73$$
Edexcel  Maths  Standard  2007  June  Core 1  Question 8
Question
8.
A sequence ??a_{1}, \: a_{2}, \: a_{3}, \: ...?? is defined by:
$$a_{1} = k,$$
$$a_{n+1} = 3a_{n} + 5, \: \: \: n \geq 1,$$
where ??k?? is a positive integer.
8. (a)
Write down an expression for ??a_{2}?? in terms of ??k??. [1]
8. (b)
Show that ??a_{3} = 9k + 20??. [2]
8. (c) (i)
Find ??\sum\limits_{r=1}^4 a_{r}?? in terms of ??k??.
8. (c) (ii)
Show that ??\sum\limits_{r=1}^4 a_{r}?? is divisble by ??10??. [4]
Hide Question
Show Solution
Worked Solution
8. (a)
Start by noting that the question asks you to write down the answer, which means it should not take very much, if any, working at all.
We have a recurrence relation here, so can simply write this for ??a_{2}?? to give the answer:
$$a_{n+1} = 3a_{n}+5$$
$$\implies a_{2} = 3a_{1} + 5$$
$$\implies a_{2} = 3k + 5$$
8. (b)
We can use the answer from 8. (a) to find ??a_{3}??:
$$a_{n+1} = 3a_{n}+5$$
$$\implies a_{3} = 3a_{2} + 5$$
$$\implies a_{3} = 3 \times (3k + 5) + 5$$
$$\implies a_{3} = 9k + 15 + 5$$
$$\implies a_{3} = 9k + 20$$
8. (c) (i)
We're being asked to calculate the sum of the terms from ??a_{1}?? to ??a_{4}?? First, we need to find ??a_{4}??:
$$a_{n+1} = 3a_{n}+5$$
$$\implies a_{4} = 3a_{3} + 5$$
$$\implies a_{4} = 3 \times (9k + 20) + 5$$
$$\implies a_{4} = 27k + 60 + 5$$
$$\implies a_{4} = 27k + 65$$
Now we can find the sum:
$$\sum\limits_{r=1}^4 a_{r}$$
$$= a_{1} + a_{2} + a_{3} + a_{4}$$
$$= k + (3k + 5) + (9k + 20) + (27k + 65) $$
$$= 40k + 90$$
8. (c) (ii)
We can rewrite the sum, by taking out a factor of ??10??:
$$\sum\limits_{r=1}^4 a_{r}$$
$$= 40k + 90$$
$$= 10(4k + 9)$$
Therefore, the sum must be divisible by ??10??.
Edexcel  Maths  Standard  2008  January  Core 1  Question 7
Question
7.
A sequence is given by:
$$x_{1} = 1,$$
$$x_{n+1} = x_{n}(p+x_{n}),$$
where ??p?? is a constant (??p \neq 0??).
7. (a)
Find ??x_{2}?? in terms of ??p??. [1]
7. (b)
Show that ??x_{3} = 1 + 3p + 2p^{2}??. [2]
Given that ??x_{3} =1??,
7. (c)
Find the value of ??p??. [3]
7. (d)
Write down the value of ??x_{2008}??. [2]
Hide Question
Show Solution
Worked Solution
7. (a)
We have a recurrence relation here, so can simply write this for ??x_{2}?? to give the answer:
$$x_{n+1} = x_{n}(p+x_{n})$$
$$\implies x_{2} = x_{1}(p + x_{1})$$
$$\implies x_{2} + 1 \times (p + 1)$$
$$\implies x_{2} = p + 1$$
7. (b)
We can use the answer from 7. (a) to find ??x_{3}??:
$$x_{n+1} = x_{n}(p+x_{n})$$
$$\implies x_{3} = x_{2}(p + x_{2})$$
$$\implies x_{3} = (p + 1) \times (p + (p +1))$$
$$\implies x_{3} = (p + 1) \times (2p + 1))$$
$$\implies x_{3} = 1 + p + 2p + 2p^{2}$$
$$\implies x_{3} = 1 + 3p + 2p^{2}$$
7. (c)
This part of the question requires the solving of a quadratic equation (albeit a simple one, using factorisation), which is from another part of the Core 1 syllabus. We can
use the statement given, that ??x_{3} = 1??, to get a quadratic in terms of ??p??:
$$x_{3} = 1 + 3p + 2p^{2}$$
$$\implies 1 = 1 + 3p + 2p^{2}$$
$$\implies 3p + 2p^{2} = 0$$
$$\implies p(3 + 2p) = 0$$
$$\implies p = 0, \: \mathrm{or} \: p = \frac{3}{2}$$
Earlier in the question it is stated that ??p \neq = 0?? therefore ??p = \frac{3}{2}??.
7. (d)
This part of the question looks tough as it is asking for the ??2008^{th}?? term. We know that ??x_{1} = 1?? and that ??x_{3}= 1??. Let's
work out a value for ??x_{2}?? (which we know an expression for from 7. (a)):
$$x_{2} = p + 1$$
$$\implies x_{2} = \frac{3}{2} + 1$$
$$\implies x_{2} = \frac{1}{2}$$
We can see, from the sequence so far ??1, \: \frac{1}{2}, \: 1, \: ...?? that subsequent terms will alternate between ??\frac{1}{2}?? and ??1??, because the recurrence relation always
stays the same. Noting that all of the evennumbered terms are ??\frac{1}{2}??
means that ??x_{2008}?? will also be ??\frac{1}{2}?? (as it is even).
Edexcel  Maths  Standard  2008  June  Core 1  Question 5
Question
5.
A sequence ??x_{1}, \: x_{2}, \: x_{3}, \: ...?? is defined by:
$$x_{1} = 1,$$
$$x_{n+1} = ax_{n}  3, \:\:\: n \geq 1,$$
where ??a?? is a constant.
5. (a)
Find an expression for ??x_{2}?? in terms of ??a??. [1]
5. (b)
Show that ??x_{3} = a^{2}  3a  3??. [2]
Given that ??x_{3} =7??,
5. (c)
Find the possible values of ??a??. [3]
Hide Question
Show Solution
Worked Solution
5. (a)
We have a recurrence relation here, so can simply write this for ??x_{2}?? to give the answer:
$$x_{n+1} = ax_{n}  3$$
$$\implies x_{2} = ax_{1}  3$$
5. (b)
We can use the answer from 4. (a) to find ??x_{3}??:
$$x_{n+1} = ax_{n}  3$$
$$\implies x_{3} = ax_{2}  3$$
$$\implies x_{3} = a \times (ax_{1}  3) 3$$
$$\implies x_{3} = a^{2}x_{1}  3a  3$$
$$\implies x_{3} = a^{2} \times 1  3a  3$$
$$\implies x_{3} = a^{2}  3a  3$$
5. (c)
This part of the question requires the solving of a quadratic equation (albeit a relatively simple one, using factorisation), which is from another part of the Core 1 syllabus. We can
use the statement given, that ??x_{3} = 7??, to get a quadratic in terms of ??a??:
$$x_{3} = a^{2}  3a  3$$
$$\implies 7 = a^{2} 3a 3$$
$$\implies a^{2}  3a  10 = 0$$
$$\implies (a  5)(a + 2) = 0$$
$$\implies a = 5, \: \mathrm{or} \: a = 2$$
Edexcel  Maths  Standard  2009  June  Core 1  Question 7
Question
7.
A sequence ??a_{1}, \: a_{2}, \: a_{3}, \: ...?? is defined by:
$$a_{1} = k,$$
$$a_{n+1} = 2a_{n}  7, \:\:\: n \geq 1,$$
where ??k?? is a constant.
7. (a)
Write down an expression for ??a_{2}?? in terms of ??k??. [1]
7. (b)
Show that ??a_{3} = 4k 21??. [2]
Given that ??\sum\limits_{r=1}^4 a_{r}=43??,
7. (c)
Find the value of ??k??. [4]
Hide Question
Show Solution
Worked Solution
7. (a)
We have a recurrence relation here, so can simply write this for ??a_{2}?? to give the answer:
$$a_{n+1} = 2a_{n}  7$$
$$\implies a_{2} = 2a_{1}  7$$
$$\implies a_{2} = 2 \times k  7$$
$$\implies a_{2} = 2k  7$$
7. (b)
We can use the answer from 7. (a) to find ??a_{3}??:
$$a_{n+1} = 2a_{n}  7$$
$$\implies a_{3} = 2a_{2}  7$$
$$\implies a_{3} = 2 \times (2k  7)  7$$
$$\implies a_{3} = 4k  14  7$$
$$\implies a_{3} = 4k  21$$
7. (c)
We should start by determining an expression for the sum of the terms from ??a_{1}?? to ??a_{4}?? First, we need to find ??a_{4}??:
$$a_{n+1} = 2a_{n}  7$$
$$\implies a_{4} = 2a_{3}  7$$
$$\implies a_{4} = 2 \times (4k  21)  7$$
$$\implies a_{4} = 8k  42  7$$
$$\implies a_{4} = 8k  49$$
We can use the statement given, that ??\sum\limits_{r=1}^4 a_{r}=43$$, to find ??k??:
$$\sum\limits_{r=1}^4 a_{r} = a_{1} + a_{2} + a_{3} + a_{4}$$
$$43 = k + (2k  7) + (4k  21) + (8k  49)$$
$$43 = 15k  77$$
$$\therefore 15k = 120$$
$$k = 8$$
Edexcel  Maths  Standard  2010  June  Core 1  Question 5
Question
5.
A sequence of positive numbers is defined by:
$$a_{n+1} = \sqrt{a_{n}^{2} + 3}, \:\:\: n \geq 1,$$
$$a_{1} = 2$$
5. (a)
Find ??a_{2}?? and ??a_{3}??, leaving your answers in surd form. [2]
5. (b)
Show that ??a_{5} = 4??. [2]
Hide Question
Show Solution
Worked Solution
5. (a)
We have a recurrence relation here, so can simply write this for ??a_{2}?? to give the first part of the answer:
$$a_{n+1} = \sqrt{a_{n}^{2} + 3}$$
$$\implies a_{2} = \sqrt{a_{1}^{2} + 3}$$
$$\implies a_{2} = \sqrt{2^{2} + 3}$$
$$\implies a_{2} = \sqrt{4 + 3}$$
$$\implies a_{2} = \sqrt{7}$$
We can now use ??a_{2}?? to find ??a_{3}?? in a similar manner:
$$a_{n+1} = \sqrt{a_{n}^{2} + 3}$$
$$\implies a_{3} = \sqrt{a_{2}^{2} + 3}$$
$$\implies a_{3} = \sqrt{(\sqrt{7})^{2} + 3}$$
$$\implies a_{3} = \sqrt{7 + 3}$$
$$\implies a_{3} = \sqrt{10}$$
5. (b)
We can use the answer from 5. (a) to find ??a_{4}??:
$$a_{n+1} = \sqrt{a_{n}^{2} + 3}$$
$$\implies a_{4} = \sqrt{a_{3}^{2} + 3}$$
$$\implies a_{4} = \sqrt{(\sqrt{10})^{2} + 3}$$
$$\implies a_{4} = \sqrt{10 + 3}$$
$$\implies a_{4} = \sqrt{13}$$
And finally, we can now use ??a_{4}?? to find ??a_{5}?? using the same methodology:
$$a_{n+1} = \sqrt{a_{n}^{2} + 3}$$
$$\implies a_{5} = \sqrt{a_{4}^{2} + 3}$$
$$\implies a_{5} = \sqrt{(\sqrt{13})^{2} + 3}$$
$$\implies a_{5} = \sqrt{13 + 3}$$
$$\implies a_{5} = \sqrt{16}$$
$$\implies a_{5} = 4$$
Edexcel  Maths  Standard  2011  January  Core 1  Question 4
Question
4.
A sequence ??a_{1}, \: a_{2}, \: a_{3}, \: ...?? is defined by:
$$a_{1} = 2,$$
$$a_{n+1} = 3a_{n}  c$$
where ??c?? is a constant.
4. (a)
Find an expression for ??a_{2}?? in terms of ??c??. [1]
Given that ??\sum\limits_{i=1}^3 a_{i}=0??
4. (b)
Find the value of ??c??. [4]
Hide Question
Show Solution
Worked Solution
4. (a)
We have a recurrence relation here, so can simply write this for ??a_{2}?? to give the answer:
$$a_{n+1} = 3a_{n}  c$$
$$\implies a_{2} = 3a_{1}  c$$
$$\implies a_{2} = 3 \times 2  c$$
$$\implies a_{2} = 6  c$$
4. (b)
We can use the answer from 4. (a) to find ??a_{3}??:
$$a_{n+1} = 3a_{n}  c$$
$$\implies a_{3} = 3a_{2}  c$$
$$\implies a_{3} = 3 \times (6  c)  c$$
$$\implies a_{3} = 18  3c  c$$
$$\implies a_{3} = 18  4c$$
We can use the statement given, that ??\sum\limits_{i=1}^3 a_{i}=0??, to find ??c??:
$$\sum\limits_{i=1}^3 a_{i} = a_{1} + a_{2} + a_{3}$$
$$0 = 2 + (6c) + (184c)$$
$$0 = 26  5c$$
$$\therefore 5c = 26$$
$$c = \frac{26}{5} ( = 5.2)$$
Edexcel  Maths  Standard  2011  June  Core 1  Question 5
Question
5.
A sequence ??a_{1}, \: a_{2}, \: a_{3}, \: ...?? is defined by:
$$a_{1} = k,$$
$$a_{n+1} = 5a_{n} + 3$$
where ??k?? is a positive integer.
5. (a)
Write down an expression for ??a_{2}?? in terms of ??k??. [1]
5. (b)
Show that ??a_{3}=25k +18??. [2]
5. (c) (i)
Find ??\sum\limits_{r=1}^4 a_{r}?? in terms of ??k??, in its simplest form.
5. (c) (ii)
Show that ??\sum\limits_{r=1}^4 a_{r}?? is divisible by ??6??. [4]
Hide Question
Show Solution
Worked Solution
5. (a)
We have a recurrence relation here, so can simply write this for ??a_{2}?? to give the answer:
$$a_{n+1} = 5a_{n} + 3$$
$$\implies a_{2} = 5a_{1} + 3$$
$$\implies a_{2} = 5k + 3$$
5. (b)
We can use the answer from 5. (a) to find ??a_{3}??:
$$a_{n+1} = 5a_{n} + 3$$
$$\implies a_{3} = 5a_{2} + 3$$
$$\implies a_{3} = 5 \times (5k + 3) + 3$$
$$\implies a_{3} = 25k + 15 + 3$$
$$\implies a_{3} = 25k + 18$$
5. (c) (i)
We're being asked to calculate the sum of the terms from ??a_{1}?? to ??a_{4}?? The easiest way of doing this is
to calculate ??a_{4}??:
$$a_{n+1} = 5a_{n} + 3$$
$$\implies a_{4} = 5a_{3} + 3$$
$$\implies a_{4} = 5 \times (25k + 18) + 3$$
$$\implies a_{4} = 125k + 90 + 3$$
$$\implies a_{4} = 125k + 93$$
Now, we can calculate the required sum:
$$\sum\limits_{r=1}^4 a_{r}$$
$$= a_{1} + a_{2} + a_{3} + a_{4}$$
$$= k + (5k + 3) + (25k + 18) + (125k + 93)$$
$$= 156k + 114$$
5. (c) (ii)
We can rewrite the sum, by taking out a factor of ??6??:
$$\sum\limits_{r=1}^4 a_{r}$$
$$= 156k + 114$$
$$= 6(26k + 19)$$
Therefore, the sum must be divisible by ??6??.
Edexcel  Maths  Standard  2012  January  Core 1  Question 4
Question
4.
A sequence ??x_{1}, \: x_{2}, \: x_{3}, \: ...?? is defined by:
$$x_{1} = 1,$$
$$x_{n+1} = ax_{n} + 4, \: \: \: n \geq 1$$
where ??a?? is a constant.
4. (a)
Write down an expression for ??x_{2}?? in terms of ??a??. [1]
4. (b)
Show that ??x_{3}=a^{2} + 5a + 5??. [2]
4. (c)
Given that ??x_{3} = 41??, find the possible values of ??a??. [3]
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4. (a)
We have a recurrence relation here, so can simply write this for ??x_{2}?? to give the answer:
$$x_{n+1} = ax_{n} + 5$$
$$\implies x_{2} = ax_{1} + 5$$
$$\implies x_{2} = a \times 1 + 5$$
$$\implies x_{2} = a + 5$$
4. (b)
We can use the answer from 4. (a) to find ??x_{3}??:
$$x_{n+1} = ax_{n} + 5$$
$$\implies x_{3} = ax_{2} + 5$$
$$\implies x_{3} = a \times (a + 5) + 5$$
$$\implies x_{3} = a^{2} + 5a + 5$$
4. (c)
We can use the answer from 4. (b) to solve this part of the question:
$$x_{3} = a^{2} + 5a + 5$$
$$\implies 41 = a^{2} + 5a + 5$$
$$\implies 0 = a^{2} + 5a  36$$
$$\implies a^{2} + 5a  36 = 0$$
$$\implies (a + 9)(a  4) = 0$$
$$\implies a = 9 \: \mathrm{or} \: a = 4$$
Edexcel  Maths  Standard  2012  June  Core 1  Question 5
Question
5.
A sequence of numbers ??a_{1}, \: a_{2}, \: a_{3}, \: ...?? is defined by:
$$a_{1} = 3,$$
$$a_{n+1} = 2a_{n} c, \: \: \: n \geq 1$$
where ??c?? is a constant.
5. (a)
Write down an expression in terms of ??c?? for ??a_{2}??. [1]
5. (b)
Show that ??a_{3}= 12  3c??. [2]
5. (c)
Given that ??\sum\limits_{i=1}^4 a_{i} \geq 23?? find the range of values of ??c??. [3]
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5. (a)
We have a recurrence relation here, so can simply write this for ??a_{2}?? to give the answer:
$$a_{n+1} = 2a_{n} c$$
$$\implies a_{2} = 2a_{1}  c$$
$$\implies a_{2} = 2 \times 3  c$$
$$\implies a_{2} = 6  c$$
5. (b)
We can use the answer from 5. (a) to find ??a_{3}??:
$$a_{n+1} = 2a_{n} c$$
$$\implies a_{3} = 2a_{2}  c$$
$$\implies a_{3} = 2 \times (6  c)  c$$
$$\implies a_{3} = 12  2c  c$$
$$\implies a_{3} = 12  3c$$
5. (c)
Let's start by finding an expression for the sum of the terms from ??a_{1}?? to ??a_{4}??. The easiest way of doing this is to calculate ??a_{4}??.
$$a_{n+1} = 2a_{n} c$$
$$\implies a_{4} = 2a_{3}  c$$
$$\implies a_{4} = 2 \times (12  3c)  c$$
$$\implies a_{4} = 24  6c  c$$
$$\implies a_{4} = 24  7c$$
We can now find an expression for ??\sum\limits_{i=1}^4 a_{i}??:
$$\sum\limits_{i=1}^4 a_{i}$$
$$= a_{1} + a_{2} + a_{3} + a_{4}$$
$$= 3 + (6  c) + (12  3c) + (24  7c)$$
$$= 45  11c$$
Finally, we know that the sum must be greater than or equal to ??23??. Let's find the value of ??c?? for which this is true:
$$\sum\limits_{r=1}^5 a_{r} \geq 23$$
$$\implies 45  11c \geq 23$$
$$\implies 45  23 \geq 11c$$
$$\implies 22 \geq 11c$$
$$\implies 2 \geq c$$
$$\implies c \leq 2$$
Edexcel  Maths  Standard  2013  January  Core 1  Question 4
Question
4.
A sequence ??u_{1}, \: u_{2}, \: u_{3}, \: ...?? satisfies:
$$u_{n+1} = 2u_{n}  1, \: \: \: n \geq 1$$
Given that ??u_{2} = 9??,
4. (a)
Find the value of ??u_{3}?? and the value of ??u_{4}??. [2]
4. (b)
Evaluate ??\sum\limits_{r=1}^4 u_{4}??. [3]
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4. (a)
We have a recurrence relation here, so can simply write this for ??u_{3}?? to give the first part of the answer:
$$u_{n+1} = 2u_{n}  1$$
$$\implies u_{3} = 2u_{2}  1$$
$$\implies u_{3} = 2 \times 9  1$$
$$\implies u_{3} = 18  1$$
$$\implies u_{3} = 17$$
We can now use the value we have just found for ??u_{3}?? to find ??u_{4}??:
$$u_{n+1} = 2u_{n}  1$$
$$\implies u_{4} = 2u_{3}  1$$
$$\implies u_{4} = 2 \times 17  1$$
$$\implies u_{4} = 34  1$$
$$\implies u_{4} = 33$$
4. (b)
We can't calculate the required sum without knowing one more term, ??u_{1}??. We can find this using the recurrence relationship in reverse:
$$u_{n+1} = 2u_{n}  1$$
$$\implies u_{2} = 2u_{1}  1$$
$$\implies 9 = 2u_{1}  1$$
$$\implies 9 + 1 = 2u_{1}$$
$$\implies 10 = 2u_{1}$$
$$\implies u_{1} = 5$$
Now we can find the value of the sum:
$$\sum\limits_{r=1}^4 u_{4}$$
$$= u_{1} + u_{2} + u_{3} + u_{4}$$
$$= 5 + 9 + 17 + 33$$
$$= 64$$
Edexcel  Maths  Standard  2013  June  Core 1  Question 4
Question
4.
A sequence of numbers ??a_{1}, \: a_{2}, \: a_{3}, \: ...?? is defined by:
$$a_{1} = 4,$$
$$a_{n+1} = k(a_{n} +2), \: \: \: n \geq 1$$
where ??k?? is a constant.
4. (a)
Find an expression for ??a_{2}?? in terms of ??k??. [1]
4. (b)
Given that ??\sum\limits_{r=1}^3 a_{i} = 2??, find the two possible values of ??k??. [6]
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Worked Solution
4. (a)
We have a recurrence relation here, so can simply write this for ??a_{2}?? to give the answer:
$$a_{n+1} = k(a_{n} + 2)$$
$$\implies a_{2} = k(a_{1} + 2)$$
$$\implies a_{2} = k(4 + 2)$$
$$\implies a_{2} = 6k$$
4. (b)
We need to find one more term to get an expression for the sum that is given, this is ??a_{3}?? which we can find using our answer to 4. (a):
$$a_{n+1} = k(a_{n} + 2)$$
$$\implies a_{3} = k(a_{2} + 2)$$
$$\implies a_{2} = k(6k + 2)$$
$$\implies a_{2} = 6k^{2} + 2k$$
Now we can find an expression for the sum:
$$\sum\limits_{r=1}^3 a_{i}$$
$$= a_{1} + a_{2} + a_{3}$$
$$= 4 + 6k + 6k^{2} + 2k$$
$$= 6k^{2} + 8k + 4$$
We know from the question that this sum is equal to ??2??, so we can find an expression for ??k??:
$$\sum\limits_{r=1}^3 a_{i} = 2$$
$$\implies 6k^{2} + 8k + 4 = 2$$
$$\implies 6k^{2} + 8k + 2 = 0$$
$$\implies 3k^{2} + 4k + 1 = 0$$
$$\implies (3k + 1)(k + 1) = 0$$
$$\implies k = \frac{1}{3} \: \mathrm{or} \: k = 1$$
Edexcel  Maths  International  2013  June  Core 1  Question 6
Question
6.
A sequence of numbers ??x_{1}, \: x_{2}, \: x_{3}, \: ...?? is defined by:
$$x_{1} = 1,$$
$$x_{n+1} = (x_{n})^{2}  kx_{n}, \: \: \: n \geq 1$$
where ??k?? is a constant, ??k \neq 0??.
6. (a)
Find an expression for ??x_{2}?? in terms of ??k??. [1]
6. (b)
Show that ??x_{3}= 1  3k + 2k^{2}??. [2]
Given also that ??x_{3} = 1??,
6. (c)
Calculate the value of ??k??. [3]
6. (d)
Hence find the value of ??\sum\limits_{n=1}^{100} x_{n}??. [3]
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Worked Solution
6. (a)
We have a recurrence relation here, so can simply write this for ??x_{2}?? to give the answer:
$$x_{n+1} = (x_{n})^{2}  kx_{n}$$
$$\implies x_{2} = (x_{1})^{2}  kx_{1}$$
$$\implies x_{2} = 1^{2}  k$$
$$\implies x_{2} = 1  k$$
6. (b)
We can find ??x_{3}?? by using the recurrence relation and our answer from 6. (a):
$$x_{n+1} = (x_{n})^{2}  kx_{n}$$
$$\implies x_{3} = (x_{2})^{2}  kx_{2}$$
$$\implies x_{3} = (1  k)^{2}  k(1  k)$$
$$\implies x_{3} = 1 \times 1 + 1 \times (k) + 1 times (k) + (k) \times (k)  k + k^{2}$$
$$\implies x_{3} = 1 k  k + k^{2}  k + k^{2}$$
$$\implies x_{3} = 1  3k + 2k^{2}$$
6. (c)
We can find the value of ??k?? now, by setting ??x_{3} = 1??:
$$x_{3} = 1$$
$$\implies 1 = 1  3k + 2k^{2}$$
$$\implies 0 =2k^{2}  3k$$
$$\implies k(2k  3) = 0$$
$$\implies k = 0 \: \mathrm{or} \: k = \frac{3}{2}$$
We know from the question that ??k \neq 0?? therfore:
$$k = \frac{3}{2}$$
6. (d)
Here, we must spot that the terms oscillate in value, because it is a recurrence relation that stays the same, if the same value appears more than once
in the sequence of terms, then the values following that value must always be the same too. Here we can see that:
$$x_{1} = 1$$
$$x_{2} = 1 k$$
$$x_{3} = 1$$
Therefore, ??x_{4} = 1k?? and ??x_{5} = 1??, etc. We can now find the sum by reasoning that we will have all of the odd terms ??x_{1}, x_{3}?? etc. being equal to ??1??
and all of the even terms ??x_{2}, x_{4}?? etc. being equal to ??1k??. Therefore the sum up to ??x_{100}?? is:
$$ 50 \times 1 + 50 \times (1  k)$$
$$= 50 + 50  50k$$
$$= 100  50k$$
$$= 100  50 \times \frac{3}{2}$$
$$= 100  75$$
$$= 25$$
Exam Tips

Exam questions will often give you a recurrence relation and ask you to find values later in the sequence, this can be done by substituting into the relation given.

Sometimes exam questions will ask you to find the range of possible values of a constant in the recurrence relation, typically you'll need to set the expression for
a term or sum of terms to some value, which will allow you to solve for the range of possible values of the constant.

You should be familiar with sum notation, as this is often used in questions. Remember, there is no calculator allowed, so you will be typically only
asked to add up a handful of terms. If you are asked to find the sum to a larger number of terms, say ??100??, then you will be expected to spot some pattern or trend
to help you do this  do not try to add up an unreasonable number of terms by hand!