Sequences, recurrence relations and sigma notation - Edexcel A-Level Maths (Paper 1) Exam Questions

Practise 17 exam-style questions on Sequences, recurrence relations and sigma notation for Edexcel A-Level Maths (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

17 questions
2–9 marks each
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17 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

3 marks

Given that
\[\sum_{r=1}^{20} u_r = 140 \quad \text{and} \quad \sum_{r=1}^{20} u_r^2 = 1250\]
find the value of
\[\sum_{r=1}^{20} (u_r - 2)^2\]

Model answer

\((u_r - 2)^2 = u_r^2 - 4u_r + 4\)
\(\displaystyle\sum_{r=1}^{20} (u_r - 2)^2 = \sum_{r=1}^{20} u_r^2 - 4\sum_{r=1}^{20} u_r + \sum_{r=1}^{20} 4\)
\(= 1250 - 4(140) + 20 \times 4\)
\(= 1250 - 560 + 80 = 770\)

Explanation

Expand the bracket and use the fact that a sum can be split term by term, with constants taken outside: \(\sum 4u_r = 4\sum u_r\). The constant term 4 is added once for each of the 20 values of \(r\), giving 80. The commonest error is to add 4 only once.

Mark scheme

1 mark
M1 Expands \((u_r - 2)^2 = u_r^2 - 4u_r + 4\) and splits the sum: \(\displaystyle\sum u_r^2 - 4\sum u_r + \sum_{r=1}^{20} 4\); or a correct final answer of 770 with no working
An answer of 770 with no working scores 3 marks.
1 mark
B1 \(\displaystyle\sum_{r=1}^{20} 4 = 80\)
A common error is \(\displaystyle\sum_{r=1}^{20} 4 = 4\), giving 694.
1 mark
A1 770
Depends on the M mark. cao.

Question 2

5 marks

A sequence has \(n\)th term
\[u_n = 3 + 2\cos\left(\dfrac{2n\pi}{3}\right), \quad n \geqslant 1\]
where the angle is in radians.

Find the values of \(u_1\), \(u_2\) and \(u_3\), and state the order of this periodic sequence.

Hence find
\[\sum_{n=1}^{100} u_n\]

Model answer

\(u_1 = 3 + 2\cos\dfrac{2\pi}{3} = 3 + 2\left(-\dfrac{1}{2}\right) = 2\)
\(u_2 = 3 + 2\cos\dfrac{4\pi}{3} = 3 + 2\left(-\dfrac{1}{2}\right) = 2\)
\(u_3 = 3 + 2\cos 2\pi = 3 + 2 = 5\)
\(u_4 = 3 + 2\cos\dfrac{8\pi}{3} = 2\), and the terms repeat every three terms: the order is 3.
\(\displaystyle\sum_{n=1}^{100} u_n = 33(2 + 2 + 5) + u_{100} = 297 + 2 = 299\)

Explanation

Use the exact values of cosine in radians: \(\cos\dfrac{2\pi}{3} = \cos\dfrac{4\pi}{3} = -\dfrac{1}{2}\) and \(\cos 2\pi = 1\), so the terms repeat 2, 2, 5 with order 3. The first 99 terms are 33 complete blocks with sum 9 each, and the 100th term is the first of a new block, 2. Common errors are working in degrees and forgetting the last term, giving 297.

Mark scheme

1 mark
M1 Evaluates at least two of \(u_1\), \(u_2\), \(u_3\) using exact values, e.g. \(\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}\); or correct final answers of 2, 2, 5, order 3 and 299 with no working
Answers of \(u_1 = 2\), \(u_2 = 2\), \(u_3 = 5\), order 3 and 299 with no working score 5 marks. A common error is a calculator in degree mode, which gives values close to 5: M0.
1 mark
A1 \(u_1 = 2\), \(u_2 = 2\), \(u_3 = 5\)
All three needed.
1 mark
B1 Order 3
Do not accept \(\pm 3\).
1 mark
M1 Uses 33 complete blocks of three terms (sum 9 each) plus \(u_{100} = u_1\): \(33 \times 9 + 2\)
Must use their block sum and deal with the one remaining term. Using an arithmetic series formula is M0.
1 mark
A1 299
cao. Common errors are 297 (the last term forgotten) and 306 (34 complete blocks).

Question 3

5 marks

A sequence has \(n\)th term
\[u_n = \dfrac{kn + 3}{n + 1}, \quad n \geqslant 1\]
where \(k\) is a constant.

Using algebra, find the set of values of \(k\) for which the sequence is decreasing.

Describe the sequence when \(k = 3\).

Model answer

\(u_{n+1} - u_n = \dfrac{k(n + 1) + 3}{n + 2} - \dfrac{kn + 3}{n + 1}\)
\(= \dfrac{(kn + k + 3)(n + 1) - (kn + 3)(n + 2)}{(n + 1)(n + 2)}\)
\(= \dfrac{kn^2 + 2kn + 3n + k + 3 - (kn^2 + 2kn + 3n + 6)}{(n + 1)(n + 2)}\)
\(= \dfrac{k - 3}{(n + 1)(n + 2)}\)
For \(n \geqslant 1\), \((n + 1)(n + 2) > 0\), so \(u_{n+1} < u_n\) for all \(n\) exactly when \(k - 3 < 0\).
The sequence is decreasing for \(k < 3\).
When \(k = 3\), \(u_n = \dfrac{3(n + 1)}{n + 1} = 3\), so every term is 3 (a constant sequence).

Explanation

A sequence is decreasing when \(u_{n+1} - u_n < 0\) for every \(n\). Over the common denominator \((n + 1)(n + 2)\), which is always positive, the numerator simplifies to \(k - 3\), so the sequence is decreasing exactly when \(k < 3\). When \(k = 3\) the numerator and denominator have the common factor \(n + 1\) and every term is 3; checking particular values is not enough when algebra is asked for.

Mark scheme

1 mark
M1 Forms \(u_{n+1} - u_n = \dfrac{k(n + 1) + 3}{n + 2} - \dfrac{kn + 3}{n + 1}\) and writes it over the common denominator \((n + 1)(n + 2)\). An answer with no working scores 0 marks for the first four marks
An answer of \(k < 3\) with no working scores 0 marks for the first four marks: the question asks for algebra. Testing particular values of \(k\) or \(n\) scores no marks. The final B1 (describing the sequence when \(k = 3\)) is marked independently.
1 mark
A1 \(u_{n+1} - u_n = \dfrac{k - 3}{(n + 1)(n + 2)}\)
The numerator \([k(n + 1) + 3](n + 1) - (kn + 3)(n + 2)\) must be expanded correctly to \(k - 3\).
1 mark
M1 States that \((n + 1)(n + 2) > 0\) for \(n \geqslant 1\), so the sign of \(u_{n+1} - u_n\) is the sign of \(k - 3\)
1 mark
A1 \(k < 3\)
Accept \(\{k : k < 3\}\). \(k \leqslant 3\) is A0.
1 mark
B1 When \(k = 3\), \(u_n = \dfrac{3n + 3}{n + 1} = 3\) for all \(n\): every term is 3 (a constant sequence, neither increasing nor decreasing)
Independent of the other marks. Accept "constant" or "all terms equal 3".

Question 4

7 marks

In a training plan, Zara swims \(d_n\) metres in her \(n\)th session, where
\[d_1 = 800, \quad d_{n+1} = 0.75d_n + 600, \quad n \geqslant 1\]
Find the number of the first session in which she swims more than 2000 metres.

The sequence \(d_n\) is increasing and converges to a limit \(L\). Find the value of \(L\) and interpret it in the context of the training plan.

Her coach advises that no session should be longer than 2.5 km. Comment on whether the training plan follows this advice.

Model answer

\(d_2 = 0.75(800) + 600 = 1200\)
\(d_3 = 0.75(1200) + 600 = 1500\)
\(d_4 = 0.75(1500) + 600 = 1725\)
\(d_5 = 0.75(1725) + 600 = 1893.75\)
\(d_6 = 0.75(1893.75) + 600 = 2020.3125\)
So the first session with more than 2000 m is session 6.
\(L = 0.75L + 600\), so \(0.25L = 600\) and \(L = 2400\)
In the long term the distance per session approaches 2400 m (2.4 km).
The distances increase towards 2400 m and never exceed it, and 2400 m < 2500 m, so no session is longer than 2.5 km: the plan follows the advice.

Explanation

Iterate the recurrence until a term is more than 2000: \(d_5 = 1893.75\) and \(d_6 = 2020.3125\). If the sequence converges to \(L\), then \(L\) satisfies \(L = 0.75L + 600\), giving 2400 m, the distance each session approaches in the long term. Because the distances increase towards 2400 m, which is below 2.5 km, the plan never breaks the coach's advice.

Mark scheme

1 mark
M1 Uses the recurrence relation to find successive terms, with at least \(d_2 = 1200\) and \(d_3 = 1500\) correct; or correct final answers of session 6 and \(L = 2400\) with no working (special case: 2 marks)
Answers of session 6 and \(L = 2400\) with no working score 2 marks (special case: the A1 for session 6 and the A1 for \(L = 2400\)); the terms and the equation for \(L\) must be seen for the other marks, and the two B1 marks need the interpretation and the comment.
1 mark
A1 \(d_5 = 1893.75\) and \(d_6 = 2020.3125\)
Both values needed. These values are exact. Accept them rounded to 1894 and 2020, i.e. [1893.5, 1894.5) and [2019.5, 2020.5).
1 mark
A1 Session 6
cao. Do not accept \(n = 6\) given with \(d_5\) only.
1 mark
M1 Sets \(L = 0.75L + 600\)
Iterating until the terms settle at 2400 is also acceptable for both marks.
1 mark
A1 \(L = 2400\)
1 mark
B1 In the long term, the distance Zara swims in each session gets closer and closer to 2400 m (2.4 km)
Must refer to the distance per session. Do not accept "she swims 2400 m in total".
1 mark
B1 The distances increase towards 2400 m but never exceed it, and 2400 m is less than 2.5 km, so the plan follows the advice
Needs a comparison of 2400 m (or 2.4 km) with 2.5 km and a conclusion.

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