Sequences, recurrence relations and sigma notation - Edexcel A-Level Maths (Paper 2) Exam Questions

Practise 16 exam-style questions on Sequences, recurrence relations and sigma notation for Edexcel A-Level Maths (Paper 2). 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.

16 questions
3–9 marks each
Instant marking
No sign-up needed

16 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

Find the value of
\[\sum_{r=11}^{30} (2r + 7)\]

[Arithmetic series: \(S_n = \dfrac{1}{2}n(a + l) = \dfrac{1}{2}n[2a + (n - 1)d]\)]

Model answer

The terms are \(2(11) + 7 = 29\), 31, 33, ..., \(2(30) + 7 = 67\): an arithmetic series with \(30 - 11 + 1 = 20\) terms.
\(\displaystyle\sum_{r=11}^{30} (2r + 7) = \dfrac{1}{2} \times 20 \times (29 + 67) = 10 \times 96 = 960\)

Explanation

The sum starts at \(r = 11\), so the first term is 29 and there are \(30 - 11 + 1 = 20\) terms; the series is arithmetic with common difference 2. Using \(\dfrac{1}{2}n(a + l)\) gives \(10 \times 96 = 960\), or subtract the first 10 terms from the first 30. The commonest error is miscounting the terms as 19.

Mark scheme

1 mark
M1 A correct method: an arithmetic series with first term 29 (from \(r = 11\)), last term 67 and 20 terms, or \(\displaystyle\sum_{r=1}^{30} - \sum_{r=1}^{10}\); or a correct final answer of 960 with no working
An answer of 960 with no working scores 3 marks. Listing all 20 terms and adding them is a valid method.
1 mark
A1 20 terms with first term 29 and last term 67, or \(\displaystyle\sum_{r=1}^{30}(2r + 7) = 1140\) and \(\displaystyle\sum_{r=1}^{10}(2r + 7) = 180\)
A common error is to use 19 terms (\(30 - 11\)) or 21 terms: A0. Using 9 (the term for \(r = 1\)) as the first term is A0.
1 mark
A1 960
Depends on the M mark. cao.

Question 2

6 marks

The sum of the first \(n\) terms of a sequence \(u_1, u_2, u_3, \dots\) is given by
\[\sum_{r=1}^{n} u_r = n^2 + 5n\]
for all positive integers \(n\).

Show that \(u_r = 2r + 4\).

Find, in terms of \(n\), \(\displaystyle\sum_{r=n+1}^{2n} u_r\), giving your answer in its simplest form.

Hence find the value of \(n\) for which \(\displaystyle\sum_{r=n+1}^{2n} u_r = 572\).

Model answer

\(u_n = \displaystyle\sum_{r=1}^{n} u_r - \sum_{r=1}^{n-1} u_r = n^2 + 5n - \left[(n - 1)^2 + 5(n - 1)\right]\)
\(= n^2 + 5n - n^2 + 2n - 1 - 5n + 5 = 2n + 4\), so \(u_r = 2r + 4\)
\(\displaystyle\sum_{r=n+1}^{2n} u_r = \sum_{r=1}^{2n} u_r - \sum_{r=1}^{n} u_r = (4n^2 + 10n) - (n^2 + 5n) = 3n^2 + 5n\)
\(3n^2 + 5n = 572\), so \(3n^2 + 5n - 572 = 0\), \((n - 13)(3n + 44) = 0\)
\(n\) is a positive integer, so \(n = 13\).

Explanation

The \(n\)th term is the difference between the sum of the first \(n\) terms and the sum of the first \(n - 1\) terms, which gives \(u_n = 2n + 4\). The sum from \(r = n + 1\) to \(2n\) is the sum of the first \(2n\) terms minus the sum of the first \(n\), so \(3n^2 + 5n\), and solving \(3n^2 + 5n = 572\) gives \(n = 13\). Common errors are subtracting the sum of the first \(n + 1\) terms instead of \(n\), and bracket errors in \((n - 1)^2 + 5(n - 1)\).

Mark scheme

1 mark
M1 Uses \(u_n = \displaystyle\sum_{r=1}^{n} u_r - \sum_{r=1}^{n-1} u_r = n^2 + 5n - \left[(n - 1)^2 + 5(n - 1)\right]\). An answer with no working scores 0 marks for the show-that.
Answers of \(3n^2 + 5n\) and \(n = 13\) with no working score 2 marks (special case: the A1 for \(3n^2 + 5n\) and the final A1); the show-that needs working. Alternative method: showing that \(\displaystyle\sum_{r=1}^{n}(2r + 4) = n(n + 1) + 4n = n^2 + 5n\) for all \(n\), with a conclusion, scores M1 A1*.
1 mark
A1* \(u_n = n^2 + 5n - (n^2 - 2n + 1) - 5n + 5 = 2n + 4\), so \(u_r = 2r + 4\)
cso. Brackets must be expanded correctly. Checking only \(u_1 = 6\) and \(u_2 = 8\) is not enough.
1 mark
M1 Uses \(\displaystyle\sum_{r=n+1}^{2n} u_r = \sum_{r=1}^{2n} u_r - \sum_{r=1}^{n} u_r = (2n)^2 + 5(2n) - (n^2 + 5n)\)
Or an arithmetic series with \(n\) terms, first term \(2(n + 1) + 4\) and last term \(2(2n) + 4\). Using \(\sum_{r=1}^{2n} - \sum_{r=1}^{n+1}\) is M0.
1 mark
A1 \(3n^2 + 5n\)
Accept \(n(3n + 5)\).
1 mark
M1 Solves \(3n^2 + 5n = 572\), a three-term quadratic
Follow through their quadratic expression; a calculator may be used.
1 mark
A1 \(n = 13\)
The negative root \(-\dfrac{44}{3}\) must be rejected or not given.

Question 3

7 marks

A sequence is defined by
\[u_n = n(0.8)^n, \quad n \geqslant 1\]

Show that \(\dfrac{u_{n+1}}{u_n} = \dfrac{4(n + 1)}{5n}\).

Hence show that the sequence is neither increasing nor decreasing, and find the exact value of the greatest term of the sequence.

Model answer

\(\dfrac{u_{n+1}}{u_n} = \dfrac{(n + 1)(0.8)^{n+1}}{n(0.8)^n} = \dfrac{0.8(n + 1)}{n} = \dfrac{4(n + 1)}{5n}\)
Every term is positive, so \(u_{n+1} > u_n\) exactly when \(\dfrac{4(n + 1)}{5n} > 1\), i.e. \(4n + 4 > 5n\), i.e. \(n < 4\).
So \(u_1 < u_2 < u_3 < u_4\); when \(n = 4\) the ratio is 1, so \(u_5 = u_4\); for \(n \geqslant 5\) the ratio is less than 1, so \(u_5 > u_6 > u_7 > \dots\)
The sequence increases and then decreases, so it is neither increasing nor decreasing.
The greatest terms are \(u_4 = 4(0.8)^4 = 4 \times \dfrac{256}{625} = \dfrac{1024}{625}\) and \(u_5 = 5(0.8)^5 = 5 \times \dfrac{1024}{3125} = \dfrac{1024}{625}\).
The greatest term is \(\dfrac{1024}{625}\).

Explanation

For a sequence of positive terms, \(u_{n+1} > u_n\) exactly when \(\dfrac{u_{n+1}}{u_n} > 1\), and here the ratio \(\dfrac{4(n + 1)}{5n}\) is greater than 1 only for \(n < 4\). So the terms rise to \(u_4\), \(u_5\) is equal to \(u_4\), and the terms then fall, giving the greatest value \(\dfrac{1024}{625}\). A common error is to stop at \(n = 4\) without noticing that \(u_5 = u_4\), or to give a rounded decimal when the exact value is asked for.

Mark scheme

1 mark
M1 Writes \(\dfrac{u_{n+1}}{u_n} = \dfrac{(n + 1)(0.8)^{n+1}}{n(0.8)^n}\) and simplifies the powers to \(\dfrac{0.8(n + 1)}{n}\). An answer with no working scores 0 marks for the show-that.
An answer of \(\dfrac{1024}{625}\) with no working scores 1 mark (special case: the final A1); the show-thats need working.
1 mark
A1* \(\dfrac{u_{n+1}}{u_n} = \dfrac{0.8(n + 1)}{n} = \dfrac{4(n + 1)}{5n}\)
cso.
1 mark
M1 Compares the ratio with 1, using the fact that every term is positive: \(\dfrac{4(n + 1)}{5n} > 1\) when \(4n + 4 > 5n\), i.e. \(n < 4\)
Or finds the ratio is greater than 1 for \(n = 1, 2, 3\), equal to 1 for \(n = 4\) and less than 1 for \(n \geqslant 5\).
1 mark
A1 \(u_{n+1} > u_n\) for \(n = 1, 2, 3\); \(u_5 = u_4\); \(u_{n+1} < u_n\) for \(n \geqslant 5\)
All three statements needed (the equality may be implied by "\(u_4 = u_5\)").
1 mark
B1 Concludes that the sequence is neither increasing nor decreasing, e.g. because \(u_2 > u_1\) but \(u_6 < u_5\)
Needs a reason. Listing terms such as 0.8, 1.28, 1.536, 1.6384, 1.6384, 1.5729 with this conclusion is acceptable for this mark.
1 mark
M1 Evaluates \(u_4\) or \(u_5\), e.g. \(4(0.8)^4 = 4 \times \dfrac{256}{625}\)
Their greatest term must come from a correct comparison of terms.
1 mark
A1 \(\dfrac{1024}{625}\) (both \(u_4\) and \(u_5\))
Accept the exact decimal 1.6384. A rounded value such as 1.64 is A0.

Question 4

8 marks

A sequence is defined by
\[u_1 = k, \quad u_{n+1} = u_n^2 - 4u_n + 6, \quad n \geqslant 1\]
where \(k\) is a constant.

Find the two values of \(k\) for which every term of the sequence is equal to \(k\).

Show that \(u_{n+1} - 2 = (u_n - 2)^2\) for all \(n \geqslant 1\).

Given that \(k = 4\), find \(u_2\) and \(u_3\), and explain why the sequence is increasing.

Given instead that \(k = 1\), describe the behaviour of the sequence.

Model answer

Every term equals \(k\) when \(u_2 = u_1\): \(k^2 - 4k + 6 = k\), so \(k^2 - 5k + 6 = 0\) and \((k - 2)(k - 3) = 0\)
\(k = 2\) or \(k = 3\)
\(u_{n+1} - 2 = u_n^2 - 4u_n + 6 - 2 = u_n^2 - 4u_n + 4 = (u_n - 2)^2\)
\(k = 4\): \(u_2 = 16 - 16 + 6 = 6\), \(u_3 = 36 - 24 + 6 = 18\)
\(u_1 - 2 = 2 > 1\). If \(u_n - 2 > 1\), then \(u_{n+1} - 2 = (u_n - 2)^2 > 1\), so every term satisfies \(u_n - 2 > 1\).
For a number \(t > 1\), \(t^2 > t\), so \(u_{n+1} - 2 = (u_n - 2)^2 > u_n - 2\), i.e. \(u_{n+1} > u_n\) for all \(n\): the sequence is increasing.
\(k = 1\): \(u_2 = 1 - 4 + 6 = 3\), and since \(3^2 - 4(3) + 6 = 3\), every later term is 3.
The sequence is 1, 3, 3, 3, ...: it is constant from the second term onwards.

Explanation

Every term equals \(k\) exactly when \(u_2 = u_1\), which gives \(k^2 - 5k + 6 = 0\), so \(k = 2\) or 3. Subtracting 2 from the recurrence gives \(u_{n+1} - 2 = (u_n - 2)^2\): when \(k = 4\), \(u_n - 2\) starts at 2 and each term squares the one before, so it stays above 1 and keeps growing, and the sequence is increasing. When \(k = 1\) the second term is 3, one of the constant values, so the sequence stays at 3; a common error is to explain "increasing" by listing a few terms only.

Mark scheme

1 mark
M1 Sets \(u_2 = u_1\): \(k^2 - 4k + 6 = k\), and solves the three-term quadratic \(k^2 - 5k + 6 = 0\); or correct final answers of \(k = 2\) and \(k = 3\) with no working
Answers of \(k = 2\) and \(k = 3\) with no working score 2 marks, and \(u_2 = 6\), \(u_3 = 18\) with no working score the B1; the show-that and the two explanations need working, so a bare answer to the whole question scores at most 3 marks. The quadratic may be solved on a calculator.
1 mark
A1 \(k = 2\) and \(k = 3\)
Both needed (from \((k - 2)(k - 3) = 0\)).
1 mark
M1 Subtracts 2 from both sides of the recurrence, \(u_{n+1} - 2 = u_n^2 - 4u_n + 4\), or expands \((u_n - 2)^2 = u_n^2 - 4u_n + 4\)
Working from both sides to meet in the middle is acceptable with a conclusion.
1 mark
A1* \(u_{n+1} - 2 = u_n^2 - 4u_n + 4 = (u_n - 2)^2\)
cso. Checking particular values of \(n\) only is A0.
1 mark
B1 \(u_2 = 6\) and \(u_3 = 18\)
Both needed. From \(16 - 16 + 6\) and \(36 - 24 + 6\).
1 mark
M1 A valid method for showing that \(u_{n+1} > u_n\) when \(k = 4\): uses \(u_{n+1} - 2 = (u_n - 2)^2\) with \(u_n - 2 > 1\), so that \((u_n - 2)^2 > u_n - 2\); or uses \(u_{n+1} - u_n = u_n^2 - 5u_n + 6 = (u_n - 2)(u_n - 3)\) with \(u_n > 3\)
The given result may be used even if it was not shown. Alternative method: writing \(v_n = u_n - 2\), so that \(v_1 = 2\) and \(v_{n+1} = v_n^2\), giving \(v_n = 2^{2^{n-1}}\), which increases with \(n\), scores M1 A1. Calculating the first few terms only is M0.
1 mark
A1 Complete explanation: \(u_1 - 2 = 2 > 1\), and whenever \(u_n - 2 > 1\) the next term has \(u_{n+1} - 2 = (u_n - 2)^2 > 1\), so every term has \(u_n - 2 > 1\) (\(u_n > 3\)); hence \(u_{n+1} - 2 > u_n - 2\), i.e. \(u_{n+1} > u_n\) for all \(n\), and the sequence is increasing
An argument that each term passes the property on to the next is enough (formal induction is not required), but the reason why every term is greater than 3 (or why \(u_n - 2 > 1\)) must be given.
1 mark
B1 \(k = 1\): \(u_2 = 1 - 4 + 6 = 3\), and then every later term is 3 (since \(3^2 - 4(3) + 6 = 3\)): the sequence 1, 3, 3, 3, ... is constant from the second term
Needs \(u_2 = 3\) and the statement that the terms then stay at 3. Do not accept "periodic" or "increasing".

Ready to try the rest? Every answer you write is marked against the mark scheme, so you can see exactly where the marks are.

Study Helper Hints first, answers after you try

Stuck on this one? Ask for a hint and I’ll point you in the right direction, without giving the answer away. Once you’ve checked your answer, I’ll show you how the marks are awarded.