4 Sequences and series — revision question pack

6 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9MA0 section 4. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

How this checking works

4.1 · Understand and use the binomial expansion of (a + bx)ⁿ for positive integer n; the notations n! and nCr; link to binomial probabilities; extend to any rational n, including for approximation, valid for |bx/a| < 1.

Explanation

For a positive integer $n$ , $(a+bx)^n=\sum_{r=0}^{n}\binom{n}{r}a^{n-r}(bx)^r$ , where $\binom{n}{r}=n!/[r!(n-r)!]$ , $n!=n(n-1)\cdots1$ and $0!=1$ ; Pascal's relation is $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$ for $1\le r\le n-1$ . The same coefficients appear in binomial probabilities: $P(X=r)=\binom{n}{r}p^r(1-p)^{n-r}$ for $X\sim B(n,p)$ .
For rational $n$ , write the expression as $a^n(1+u)^n$ and use $1+nu+n(n-1)u^2/2!+\cdots$ , valid for $|u|<1$ .
For an approximation, choose a nearby convenient value and retain the requested number of terms.
A common error is to omit powers of the coefficient $b$ from terms involving $(bx)^r$ .
Use $(1+u)^n=1+nu+n(n-1)u^2/2+\cdots$ with $n=-1/2$ and $u=-2x$ .

Worked example

For $(1-2x)^{-1/2}$ , obtain the constant, $x$ and $x^2$ terms. Also give the interval of $x$ on which this series is valid.

1.Use $(1+u)^n=1+nu+n(n-1)u^2/2+\cdots$ with $n=-1/2$ and $u=-2x$ .
2.This gives $1+(-1/2)(-2x)+[(-1/2)(-3/2)/2](-2x)^2=1+x+3x^2/2$ .
3.Validity requires $|-2x|<1$ , so $|x|<1/2$ .

Answer: $1+x+\dfrac{3}{2}x^2$ ; $|x|<\dfrac{1}{2}$

Common mistakes

Don't expand $(a+bx)^n$ without first factoring out $a$ , so the generalised binomial coefficients are applied to the wrong expression.
Don't use the finite binomial formula for a negative or fractional power and omit the convergence condition.

Exam tip

For a rational-power expansion, factor the constant first and state the validity condition from the transformed bracket.

Tier 1 · Easy

1.
Find the coefficient of $x^2$ in the expansion of $(1+4x)^5$ .
(2)
(Total for Question 1 is 2 marks)
2.
Expand $(3-2x)^4$ up to and including the term in $x^2$ . Write the result in ascending powers of $x$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Find the coefficient of $x^3$ in the expansion of $(2-x)^7$ .
(3)
(Total for Question 1 is 3 marks)
2.
Expand $(1+3x)^{-2}$ up to and including the term in $x^3$ , giving the terms in ascending powers of $x$ . State the range of $x$ -values for which the expansion is valid.
(5)
(Total for Question 2 is 5 marks)
3.
Find the expansion of $\dfrac{(1+2x)^{1/2}}{1-x}$ through the term in $x^2$ , arranging the terms by ascending powers of $x$ . State the set of $x$ -values for which both component expansions are valid.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
Use the first three terms of a binomial expansion of $(64+x)^{1/3}$ to estimate $\sqrt[3]{65}$ . Give the estimate to $5$ decimal places and justify that the expansion is valid at the value of $x$ used.
(5)
(Total for Question 1 is 5 marks)
2.
In the expansion of $(1+px)(2-x)^6$ , the term in $x^2$ vanishes, where $p$ is a constant. Find $p$ and hence determine the coefficient of $x^3$ .
(5)
(Total for Question 2 is 5 marks)
3.
Let $X$ have the binomial distribution $B(8,p)$ , where $0<p<1$ . In the expansion of $(1-p+px)^8$ , the coefficient of $x^4$ is $10$ times the coefficient of $x^2$ . Find $p$ . Hence find the exact value of $P(X\geq7)$ .
(5)
(Total for Question 3 is 5 marks)
4.
The expansion of $(1+kx)^p$ begins $1+6x+3x^2+\cdots$ , where $k>0$ and $p$ is rational. Find $p$ and $k$ . Hence find the coefficient of $x^3$ and state the set of $x$ -values for which the expansion is valid.
(6)
(Total for Question 4 is 6 marks)
5.
Use the first three terms of the binomial expansion of $(1+u)^{1/2}$ to obtain a rational estimate for $\sqrt{63}$ . State why the chosen value of $u$ meets the validity condition. By comparing the square of the estimate with $63$ , determine whether it is an overestimate or an underestimate.
(6)
(Total for Question 5 is 6 marks)

4.2 · Work with sequences including those given by a formula for the nth term and those generated by a simple relation of the form xₙ₊₁ = f(xₙ); increasing sequences; decreasing sequences; periodic sequences.

Explanation

An explicit rule gives $x_n$ directly from $n$ , whereas a recurrence relation defines each term from one or more preceding terms and needs an initial value.
A sequence is increasing when $x_{n+1}>x_n$ and decreasing when $x_{n+1}<x_n$ throughout the stated range of $n$ .
A sequence is periodic if its terms repeat after a fixed positive number of steps; the least such number is its period.
When using a recurrence, retain sufficient accuracy between steps and check any invariant interval.
A common error is to apply $f$ repeatedly to the initial term instead of to the latest term.

Worked example

A sequence is defined by $x_1=1$ and $x_{n+1}=5-x_n$ . Find $x_2,x_3,x_4,x_5$ and state its period.

1.Apply the recurrence to the most recent term: $x_2=5-1=4$ , $x_3=5-4=1$ , and the pair $1,4$ then repeats.
2.Thus $x_4=4$ , $x_5=1$ , and the least period is $2$ .

Answer: $x_2=4$ , $x_3=1$ , $x_4=4$ , $x_5=1$ ; Period $2$

Common mistakes

Don't declare a sequence increasing because $x_2>x_1$ without checking later terms or the recurrence.
Don't classify a recurrence from too few terms and miss a repeating cycle.

Exam tip

Generate enough consecutive terms to demonstrate the claimed monotonicity or period rather than guessing from the first change.

Tier 1 · Easy

1.
The sequence $(x_n)$ is defined by $x_n=7-3n$ for $n\ge1$ . Write down its first three terms and state whether it is increasing or decreasing.
(2)
(Total for Question 1 is 2 marks)
2.
A sequence is defined by $x_1=14$ and $x_{n+1}=\dfrac12x_n-1$ . Find $x_2$ , $x_3$ and $x_4$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
The sequence is defined by $x_n=\dfrac{n}{n+1}$ for $n\geq1$ . Prove that the sequence is increasing.
(4)
(Total for Question 1 is 4 marks)
2.
For $n\geq1$ , let $x_n=n^2-kn$ , where $k$ is real. Determine all values of $k$ that make $(x_n)$ increasing at every positive integer $n$ .
(4)
(Total for Question 2 is 4 marks)
3.
A sequence is defined by $x_1=k$ and $x_{n+1}=\dfrac{x_n^2+12}{7}$ , where $k$ is real. Find every value of $k$ for which the sequence is constant.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A sequence is defined by $x_1=2$ and $x_{n+1}=(x_n+6)/2$ . Let $y_n=6-x_n$ . Show that $(y_n)$ is geometric, find a formula for $x_n$ , and hence show that $(x_n)$ is increasing.
(5)
(Total for Question 1 is 5 marks)
2.
For each positive integer $n$ , let $x_n=n+\dfrac{36}{n}$ . Prove that the terms decrease as far as $x_6$ and increase thereafter. Hence identify the least term.
(5)
(Total for Question 2 is 5 marks)
3.
A sequence is defined by $x_1=6$ and $x_{n+1}=3+\dfrac{2}{x_n}$ . Find the exact values of $x_2,x_3,\ldots,x_6$ . Find both fixed points of the recurrence. By comparing only the terms you have calculated with the positive fixed point, describe their behaviour.
(5)
(Total for Question 3 is 5 marks)
4.
A sequence is defined by $x_1=2$ and $x_{n+1}=\dfrac{1}{1-x_n}$ . Calculate $x_2,x_3,x_4$ . Show algebraically that applying the recurrence three times returns any admissible term to its original value, and state the least period of this sequence. Hence find $x_{2026}$ and the exact value of $\displaystyle\sum_{r=1}^{2025}x_r$ .
(7)
(Total for Question 4 is 7 marks)
5.
For positive integers $n$ , a sequence is defined by $x_n=2\cos\left(\dfrac{n\pi}{3}\right)+(-1)^n$ . Write down $x_1,x_2,\ldots,x_6$ . Determine the least period of the sequence, justifying why no smaller positive period works. State all values taken by the sequence and describe exactly which positive integers $n$ give each value. Hence decide whether the sequence is increasing, decreasing or neither.
(5)
(Total for Question 5 is 5 marks)

4.3 · Understand and use sigma notation for sums of series.

Explanation

The notation $\sum_{r=a}^{b}f(r)$ means add the values of $f(r)$ for each integer $r$ from $a$ to $b$ inclusive.
To write a series in sigma notation, identify a formula for its general term together with the correct first and last index values.
Use linearity to split sums: for example, $\sum_{r=1}^{n}(ar+b)=a\sum_{r=1}^{n}r+b\sum_{r=1}^{n}1$ , with $\sum_{r=1}^{n}1=n$ .
Check the endpoint terms after changing an index.
A common error is to treat the upper limit as the number of terms when the lower limit is not $1$ .

Worked example

Write the series $5+9+13+\cdots+41$ using sigma notation.

1.The terms have common difference $4$ , and $4r+1$ gives $5$ when $r=1$ .
2.Solving $4r+1=41$ gives $r=10$ , so the series is $\sum_{r=1}^{10}(4r+1)$ .

Answer: $\sum_{r=1}^{10}(4r+1)$

Common mistakes

Don't shift the index but leave the summand unchanged, so the re-indexed sum represents different terms.
Don't write a sigma expression whose lower and upper limits produce the wrong first or last term.

Exam tip

Check a sigma answer by substituting both endpoint indices and confirming the number of generated terms.

Tier 1 · Easy

1.
Evaluate $\sum_{r=1}^{4}(2r+1)$ .
(2)
(Total for Question 1 is 2 marks)
2.
Write the series $11+16+21+\cdots+71$ in the form $\displaystyle\sum_{r=1}^{n}(ar+b)$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Evaluate $\displaystyle\sum_{r=1}^{8}(3r^2-2r)$ .
(3)
(Total for Question 1 is 3 marks)
2.
Rewrite $\displaystyle\sum_{r=4}^{12}(7-2r)$ as a sum whose lower limit is $1$ . Hence evaluate the sum.
(4)
(Total for Question 2 is 4 marks)
3.
Given that $\displaystyle\sum_{r=1}^{7}a_r=23$ and $\displaystyle\sum_{r=1}^{7}b_r=-4$ , evaluate $\displaystyle\sum_{r=1}^{7}(2a_r-3b_r+1)$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
Given that $\sum_{r=1}^{n}(4r-1)=210$ , find the positive integer $n$ . Show all stages of your working.
(5)
(Total for Question 1 is 5 marks)
2.
Given that $\displaystyle\sum_{r=k}^{2k}(3r-1)=85$ , where $k$ is a positive integer, find $k$ . Show all stages of your working.
(5)
(Total for Question 2 is 5 marks)
3.
Constants $a$ and $b$ satisfy the identity $\displaystyle\sum_{r=1}^{n}(ar+b)=n(2n+5)$ for every positive integer $n$ . Find $a$ and $b$ , showing how the identity determines both constants.
(4)
(Total for Question 3 is 4 marks)
4.
For a positive integer $n$ , let $S_n=\displaystyle\sum_{r=1}^{2n}(-1)^r(3r-1)$ . By grouping consecutive pairs of terms, show that $S_n=3n$ . Obtain a corresponding expression for $\displaystyle\sum_{r=1}^{2n+1}(-1)^r(3r-1)$ . Hence find $n$ when this second sum is $-44$ .
(5)
(Total for Question 4 is 5 marks)
5.
The positive integer $n$ satisfies $\displaystyle\sum_{r=1}^{n}|2r-7|=58$ . Find $n$ , showing how the modulus affects the sum.
(5)
(Total for Question 5 is 5 marks)

4.4 · Understand and work with arithmetic sequences and series, including the formulae for nth term and the sum to n terms.

Explanation

An arithmetic sequence has constant difference $d$ and nth term $u_n=a+(n-1)d$ , where $a$ is the first term.
The sum of its first $n$ terms is $S_n=n[2a+(n-1)d]/2$ , equivalently $S_n=n(a+l)/2$ when the last term $l$ is known.
Pairing the first and last terms, then the second and penultimate terms, gives $n$ equal pairs of total $a+l$ across two copies of the series, proving $S_n=n(a+l)/2$ .
The term number $n$ must be a positive integer.
A common error is to use $a+nd$ for the nth term, shifting every term by one difference.

Worked example

An arithmetic series has first term $7$ , last term $151$ and common difference $8$ . Find the number of terms and the sum of the series.

1.Use $151=7+(n-1)8$ , giving $144=8(n-1)$ and $n=19$ .
2.Then $S_{19}=19(7+151)/2=19\times79=1501$ .

Answer: $19$ terms; $1501$

Common mistakes

Don't solve $u_n=L$ and round a non-integer term number instead of checking whether $L$ is actually a term.
Don't use the last term as the number of terms in the arithmetic-series formula.

Exam tip

Find the integer term count from the nth-term formula before applying the arithmetic sum formula.

Tier 1 · Easy

1.
Find the $18$ th term of the arithmetic sequence $11,17,23,\ldots$ .
(2)
(Total for Question 1 is 2 marks)
2.
The seventh term of an arithmetic sequence is $29$ and the common difference is $4$ . Find the first term and the fifteenth term.
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The fifth term of an arithmetic sequence is $18$ and the twelfth term is $46$ . Find the first term, the common difference and the sum of the first $20$ terms.
(5)
(Total for Question 1 is 5 marks)
2.
An arithmetic sequence has partial sum $S_n=3n^2+2n$ . Derive its $n$ th term and hence determine the position of the term equal to $167$ .
(4)
(Total for Question 2 is 4 marks)
3.
Five numbers are inserted between $7$ and $49$ so that all seven numbers form an arithmetic sequence. Find the five inserted numbers and their sum.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
An arithmetic sequence has first term $a$ and common difference $d$ . Its $8$ th term is $31$ and the sum of its first $20$ terms is $730$ . Find $a$ and $d$ , and find the least value of $n$ for which the nth term exceeds $100$ .
(5)
(Total for Question 1 is 5 marks)
2.
An arithmetic sequence has positive first term $a$ and positive common difference $d$ . Its first, fifth and ninth terms are the side lengths, in increasing order, of a right-angled triangle with perimeter $72$ . Find $a$ and $d$ , and then calculate $S_{20}$ .
(6)
(Total for Question 2 is 6 marks)
3.
An arithmetic sequence has first term $30$ and common difference $-4$ . Let $S_n$ denote the sum of its first $n$ terms. Determine the positive integer $n$ for which $S_n$ is greatest and find this greatest value. Find also all positive integers $n$ for which $S_n\geq96$ .
(5)
(Total for Question 3 is 5 marks)
4.
The arithmetic sequences $7,12,17,\ldots$ and $4,11,18,\ldots$ have some terms in common. Find the first four common terms. Hence find the sum of all their common terms that are less than $500$ .
(6)
(Total for Question 4 is 6 marks)
5.
The digits of a three-digit positive integer form an increasing arithmetic sequence. The integer is $15$ times the sum of its digits. Find the integer and prove that it is the only possibility.
(5)
(Total for Question 5 is 5 marks)

4.5 · Understand and work with geometric sequences and series, including the formulae for the nth term and the sum of a finite geometric series; the sum to infinity of a convergent geometric series, including the use of |r| < 1; modulus notation.

Explanation

A geometric sequence has constant ratio $r$ and nth term $u_n=ar^{n-1}$ , where $a$ is the first term.
For $r\ne1$ , subtracting $rS_n$ from $S_n$ proves $S_n=a(1-r^n)/(1-r)$ ; an equivalent form may be more convenient when $r>1$ .
A geometric series has a finite sum to infinity only when $|r|<1$ , in which case $S_\infty=a/(1-r)$ ; for a finite-sum threshold, isolate $r^n$ and use logarithms.
A negative ratio makes term signs alternate, but convergence still depends on its modulus.
A common error is to use $r<1$ instead of $|r|<1$ , which would wrongly accept ratios below $-1$ .

Worked example

Find the exact sum to infinity of $12-3+0.75-\cdots$ .

1.The common ratio is $r=-3/12=-1/4$ , and $|r|=1/4<1$ , so the series converges.
2.Its sum is $S_\infty=12/[1-(-1/4)]=12/(5/4)=48/5$ .

Answer: $\dfrac{48}{5}$

Common mistakes

Don't use $ar^n$ rather than $ar^{n-1}$ for the nth term of a geometric sequence.
Don't apply the sum-to-infinity formula without checking that the common ratio has modulus less than one.

Exam tip

State the common ratio and verify convergence before calculating an infinite geometric sum.

Tier 1 · Easy

1.
Find the $8$ th term of the geometric sequence $3,6,12,\ldots$ .
(2)
(Total for Question 1 is 2 marks)
2.
The three positive numbers $12$ , $k$ and $27$ , in that order, are consecutive terms of a geometric sequence. Find $k$ .
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A geometric sequence has positive common ratio. Its third term is $18$ and its sixth term is $486$ . Find the first term, the common ratio and the sum of the first five terms.
(5)
(Total for Question 1 is 5 marks)
2.
A convergent geometric series satisfies $S_\infty=20$ and $S_2=\dfrac{84}{5}$ . Its ratio $r$ is positive. Determine $r$ and the initial term $a$ .
(4)
(Total for Question 2 is 4 marks)
3.
Express the recurring decimal $0.234234234\ldots$ as a fraction in its simplest form by modelling the repeating blocks as a geometric series.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
A geometric series has first term $800$ and common ratio $0.9$ . Find the least value of $n$ for which the sum of the first $n$ terms exceeds $7000$ .
(5)
(Total for Question 1 is 5 marks)
2.
A convergent geometric series has positive first term $a$ and common ratio $r$ . The sum of its first two terms is $12$ , and the sum of its first four terms is $15$ . Find all possible pairs $(a,r)$ and show that both series have the same sum to infinity.
(6)
(Total for Question 2 is 6 marks)
3.
The infinite geometric series is $(2x-1)+(2x-1)\left(\dfrac{x}{3}\right)+(2x-1)\left(\dfrac{x}{3}\right)^2+\cdots$ . Find the set of real values of $x$ for which the series converges and its sum lies strictly between $4$ and $10$ .
(5)
(Total for Question 3 is 5 marks)
4.
For the first six terms of a geometric sequence, the sum of the terms in odd-numbered positions is $42$ and the sum of the terms in even-numbered positions is $21$ . Find the first term and the common ratio. Hence find the sum of the first six terms and the sum to infinity.
(6)
(Total for Question 4 is 6 marks)
5.
For a real constant $k$ , consider the geometric series with first term $5$ and common ratio $q=k^2-4k+2$ . Determine all values of $k$ for which the series converges, and find its sum in terms of $k$ on this set. Classify what happens at every boundary value of the convergence set, and explain why the series diverges for all remaining values of $k$ .
(6)
(Total for Question 5 is 6 marks)

4.6 · Use sequences and series in modelling.

Explanation

A constant additive change suggests an arithmetic model, while a constant multiplier or percentage change suggests a geometric model.
State what the term number represents and whether the initial value is $u_1$ or $u_0$ before forming a term or sum.
Repeated deposits, withdrawals or other fixed adjustments can be represented by a recurrence and often rewritten using a finite geometric sum.
Interpret results within the context and identify assumptions such as fixed rates or indefinite continuation.
A common error is to use an infinite sum when the process has not converged or has a finite stopping point.

Worked example

A ball is dropped from a height of $12$ m. After each impact it rebounds to $65\%$ of its previous maximum height. Find the total vertical distance travelled before it comes to rest, giving your answer to $3$ significant figures.

1.The initial downward distance is $12$ m.
2.The rebound heights form a geometric series with first term $12(0.65)=7.8$ and ratio $0.65$ .
3.Each rebound height is travelled once upwards and once downwards, so the total is $12+2[7.8/(1-0.65)]=56.5714\ldots$ m.
4.To $3$ significant figures this is $56.6$ m.

Answer: $56.6$ m

Common mistakes

Don't model a fixed percentage change with an additive arithmetic sequence instead of a multiplicative ratio.
Don't count the initial drop twice or omit either the upward or downward part of each rebound.

Exam tip

Build the distance model by separating the first drop from the two-way geometric rebound distances.

Tier 1 · Easy

1.
Mina saves $\pounds 25$ in the first month and increases the amount saved by $\pounds 5$ each month. Find the total she saves in the first $6$ months.
(2)
(Total for Question 1 is 2 marks)
2.
A tank initially contains $500$ litres of water. A model assumes that it loses $8\%$ of its contents each hour. Give the predicted volume after $3$ hours to $3$ significant figures.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
A theatre has $25$ rows of seats. The first row has $16$ seats and each successive row has $2$ more seats than the preceding row. Find the total number of seats.
(3)
(Total for Question 1 is 3 marks)
2.
A mosaic is built in numbered frames. Frame $1$ uses $18$ tiles, and each new frame uses $3$ more tiles than the preceding frame. There are $1500$ tiles available. Find the greatest number of complete frames that can be built and the number of tiles left unused.
(4)
(Total for Question 2 is 4 marks)
3.
The temperature $T_n$ degrees Celsius of a machine after $n$ ten-minute intervals is modelled by $T_0=84$ and $T_{n+1}=0.75T_n+5$ . Find $T_3$ and determine the steady temperature predicted by the model. State one limitation of the model.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
An account initially contains $\pounds 5000$ . At the end of each year, after interest of $3\%$ has been added, $\pounds 400$ is withdrawn. The model is $B_0=5000$ and $B_{n+1}=1.03B_n-400$ . Derive a formula for $B_n$ and find the first value of $n$ for which the model predicts a negative balance. State one limitation of the model.
(6)
(Total for Question 1 is 6 marks)
2.
Sponsor A contributes $\pounds 200$ in the first year and increases its annual contribution by $\pounds 25$ each year. Sponsor B contributes $\pounds 400$ in the first year and then contributes $90\%$ of the previous year's amount in each later year. Find the percentage of the combined total contribution over the first $8$ years that comes from Sponsor B. Give your answer to $3$ significant figures. Find also the limiting total contributed by Sponsor B.
(6)
(Total for Question 2 is 6 marks)
3.
Stage $1$ of a cable display has $6$ segments, each $9$ metres long. To create the next stage, every segment is replaced by two segments, each one third as long. The displays for different stages are built separately. Find the number of segments and the total cable length in stage $n$ . Hence find the total cable used for the first $5$ displays and the limiting total if the process continues indefinitely. State one limitation of the model.
(6)
(Total for Question 3 is 6 marks)
4.
The concentration $C_n$ milligrams per litre of a pollutant $n$ days after monitoring begins is modelled by $C_n=ar^n$ , where $a>0$ and $0<r<1$ . Measurements give $C_2=72$ and $C_5=9$ . Find $a$ and $r$ . Find the first integer value of $n$ for which the model predicts $C_n<1$ . State one limitation of the model.
(6)
(Total for Question 4 is 6 marks)
5.
A greenhouse model records a daytime temperature $D_n$ and the following night-time temperature $N_n$ , in degrees Celsius. It uses $D_1=20$ , $N_n=0.6D_n+2$ and $D_{n+1}=0.5N_n+9$ . Show that $D_{n+1}=0.3D_n+10$ . Find an exact formula for $D_n$ . Hence find the limiting daytime and night-time temperatures predicted by the model. State one limitation of the model.
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

4.1 · Understand and use the binomial expansion of (a + bx)ⁿ for positive integer n; the notations n! and nCr; link to binomial probabilities; extend to any rational n, including for approximation, valid for |bx/a| < 1.

Tier 1 · Easy

Mark scheme for 4.1 Tier 1 · Easy
Question	Scheme	Marks
1	$160$	2
(2 marks)		2
Notes
The $x^2$ term is obtained with $r=2$ : $\binom{5}{2}(4x)^2=10\times16x^2=160x^2$ . Hence the coefficient is $160$ .
2	$81-216x+216x^2$	3
(3 marks)		3
Notes
The required terms are $3^4+\binom41 3^3(-2x)+\binom42 3^2(-2x)^2$ . These simplify to $81-216x+216x^2$ .

Tier 2 · Standard

Mark scheme for 4.1 Tier 2 · Standard
Question	Scheme	Marks
1	$-560$	3
(3 marks)		3
Notes
The $x^3$ term is obtained by choosing three factors of $-x$ : $\binom{7}{3}2^{7-3}(-x)^3=35\times16\times(-x^3)=-560x^3$ . Hence the coefficient is $-560$ .
2	$1-6x+27x^2-108x^3$ $\|x\|<\dfrac13$	5
(5 marks)		5
Notes
Using $(1+u)^{-2}=1-2u+3u^2-4u^3+\cdots$ with $u=3x$ gives $1-6x+27x^2-108x^3+\cdots$ . Validity requires $\|3x\|<1$ , so $\|x\|<1/3$ .
3	$1+2x+\dfrac32x^2$ $-\dfrac12<x<\dfrac12$	5
(5 marks)		5
Notes
Use $(1+2x)^{1/2}=1+x-\dfrac12x^2+\cdots$ and $(1-x)^{-1}=1+x+x^2+\cdots$ . Multiplying and collecting terms through $x^2$ gives $1+2x+\dfrac32x^2$ . The first expansion requires $\|2x\|<1$ and the second requires $\|x\|<1$ , so both are valid when $-1/2<x<1/2$ .

Tier 3 · Hard

Mark scheme for 4.1 Tier 3 · Hard
Question	Scheme	Marks
1	$\sqrt[3]{65}\approx4.02072$ Valid because $\|1/64\|<1$	5
(5 marks)		5
Notes
Write $(64+x)^{1/3}=4(1+x/64)^{1/3}$ . Using $n=1/3$ , the first three terms give $4[1+x/192-x^2/36864]=4+x/48-x^2/9216$ . Set $x=1$ to estimate $\sqrt[3]{65}$ : $4+1/48-1/9216=4.0207248\ldots$ , so the estimate is $4.02072$ . The expansion requires $\|x/64\|<1$ , which holds when $x=1$ .
2	$p=\dfrac54$ Coefficient of $x^3$ is $140$	5
(5 marks)		5
Notes
In $(2-x)^6$ , the coefficients of $x$ , $x^2$ and $x^3$ are $-192$ , $240$ and $-160$ respectively. The $x^2$ coefficient in the product is therefore $240-192p$ . Setting this equal to zero gives $p=5/4$ . The $x^3$ coefficient is then $-160+p(240)=-160+300=140$ .
3	$p=\dfrac23$ $P(X\geq7)=\dfrac{1280}{6561}$	5
(5 marks)		5
Notes
The coefficients of $x^4$ and $x^2$ are $\binom84p^4(1-p)^4$ and $\binom82p^2(1-p)^6$ . Their ratio is $\dfrac52[p/(1-p)]^2$ . Equating this to $10$ gives $[p/(1-p)]^2=4$ . Since $0<p<1$ , $p/(1-p)>0$ , so $p/(1-p)=2$ and $p=2/3$ . Hence $P(X\geq7)=\binom87(2/3)^7(1/3)+(2/3)^8=1280/6561$ .
4	$p=\dfrac65$ , $k=5$ Coefficient of $x^3$ is $-4$ $\|x\|<\dfrac15$	6
(6 marks)		6
Notes
The coefficients of $x$ and $x^2$ give $pk=6$ and $p(p-1)k^2/2=3$ . Since $k=6/p$ , substitution in the second equation gives $18(p-1)/p=3$ , so $p=6/5$ and $k=5$ . The coefficient of $x^3$ is $[p(p-1)(p-2)/6]k^3=[(6/5)(1/5)(-4/5)/6]125=-4$ . The generalised binomial expansion requires $\|kx\|<1$ , hence $\|x\|<1/5$ .
5	$\sqrt{63}\approx\dfrac{32511}{4096}$ Valid because $\|u\|=\dfrac{1}{64}<1$ It is an overestimate	6
(6 marks)		6
Notes
Write $\sqrt{63}=8(1-1/64)^{1/2}$ , so $u=-1/64$ . Using $(1+u)^{1/2}=1+u/2-u^2/8+\cdots$ gives $8[1-1/128-1/32768]=32511/4096$ . The expansion is valid because $\|u\|=1/64<1$ . Also $(32511/4096)^2-63=513/16777216>0$ . Since the estimate is positive, its square being greater than $63$ shows that it is an overestimate of $\sqrt{63}$ .

4.2 · Work with sequences including those given by a formula for the nth term and those generated by a simple relation of the form xₙ₊₁ = f(xₙ); increasing sequences; decreasing sequences; periodic sequences.

Tier 1 · Easy

Mark scheme for 4.2 Tier 1 · Easy
Question	Scheme	Marks
1	$4,1,-2$ Decreasing	2
(2 marks)		2
Notes
Substitute $n=1,2,3$ to obtain $4,1,-2$ . Also $x_{n+1}-x_n=-3<0$ , so the sequence is decreasing.
2	$x_2=6$ , $x_3=2$ , $x_4=0$	2
(2 marks)		2
Notes
Apply the recurrence to the latest term each time: $x_2=14/2-1=6$ , $x_3=6/2-1=2$ , and $x_4=2/2-1=0$ .

Tier 2 · Standard

Mark scheme for 4.2 Tier 2 · Standard
Question	Scheme	Marks
1	$x_{n+1}-x_n=\dfrac{1}{(n+1)(n+2)}>0$ , so $(x_n)$ is increasing.	4
(4 marks)		4
Notes
Calculate the difference between consecutive terms: $x_{n+1}-x_n=\dfrac{n+1}{n+2}-\dfrac{n}{n+1}=\dfrac{(n+1)^2-n(n+2)}{(n+1)(n+2)}=\dfrac{1}{(n+1)(n+2)}$ . For every $n\geq1$ the denominator is positive, so $x_{n+1}>x_n$ and the sequence is increasing.
2	$k<3$	4
(4 marks)		4
Notes
$x_{n+1}-x_n=(n+1)^2-k(n+1)-(n^2-kn)=2n+1-k$ . This expression is smallest when $n=1$ . It is positive for every $n\geq1$ precisely when $3-k>0$ , so $k<3$ .
3	$k=3$ or $k=4$	4
(4 marks)		4
Notes
For the sequence to be constant from its first term, $x_2=x_1=k$ . Hence $(k^2+12)/7=k$ , so $k^2-7k+12=0=(k-3)(k-4)$ . Thus $k=3$ or $k=4$ . Substituting either value into the recurrence returns the same value, so each produces a constant sequence.

Tier 3 · Hard

Mark scheme for 4.2 Tier 3 · Hard
Question	Scheme	Marks
1	$y_n=4\left(\dfrac{1}{2}\right)^{n-1}$ $x_n=6-4\left(\dfrac{1}{2}\right)^{n-1}$ $(x_n)$ is increasing	5
(5 marks)		5
Notes
Using the recurrence, $y_{n+1}=6-x_{n+1}=6-(x_n+6)/2=(6-x_n)/2=y_n/2$ . Also $y_1=6-2=4$ , so $(y_n)$ is geometric with ratio $1/2$ and $y_n=4(1/2)^{n-1}$ . Therefore $x_n=6-4(1/2)^{n-1}$ . Since $(1/2)^{n-1}$ is positive and decreases as $n$ increases, subtracting it from $6$ shows that $(x_n)$ is increasing.
2	The least term is $x_6=12$	5
(5 marks)		5
Notes
$x_{n+1}-x_n=1+36/(n+1)-36/n=1-36/[n(n+1)]$ . This is negative when $n(n+1)<36$ , which holds for $n=1,\ldots,5$ , so $x_1>x_2>\cdots>x_6$ . It is positive for every $n\geq6$ because then $n(n+1)\geq42>36$ , so the sequence increases after $x_6$ . Therefore the least term is $x_6=6+36/6=12$ .
3	$x_2=\dfrac{10}{3}$ , $x_3=\dfrac{18}{5}$ , $x_4=\dfrac{32}{9}$ , $x_5=\dfrac{57}{16}$ , $x_6=\dfrac{203}{57}$ Fixed points $\dfrac{3+\sqrt{17}}{2}$ and $\dfrac{3-\sqrt{17}}{2}$ The calculated terms alternate on either side of the positive fixed point and move closer to it.	5
(5 marks)		5
Notes
Repeated substitution gives $x_2=10/3$ , $x_3=18/5$ , $x_4=32/9$ , $x_5=57/16$ and $x_6=203/57$ . A fixed point $L$ satisfies $L=3+2/L$ , so $L^2-3L-2=0$ and $L=(3\pm\sqrt{17})/2$ . The positive fixed point is approximately $3.56155$ . The listed terms are approximately $6,3.33333,3.60000,3.55556,3.56250,3.56140$ , so they lie alternately above and below that value and their observed distances from it decrease.
4	$x_2=-1$ , $x_3=\dfrac12$ , $x_4=2$ Least period $3$ $x_{2026}=2$ $\displaystyle\sum_{r=1}^{2025}x_r=\dfrac{2025}{2}$	7
(7 marks)		7
Notes
Direct substitution gives $x_2=-1$ , $x_3=1/2$ and $x_4=2$ . If $f(x)=1/(1-x)$ , then $f(f(x))=(x-1)/x$ and $f(f(f(x)))=x$ whenever the expressions are defined. Since the three values $2,-1,1/2$ are distinct, the least period is $3$ . Since $2026$ leaves remainder $1$ when divided by $3$ , $x_{2026}=x_1=2$ . Each complete block sums to $2-1+1/2=3/2$ , and $2025=675(3)$ , so the required sum is $675(3/2)=2025/2$ .
5	$0,0,-3,0,0,3$ Least period $6$ $x_n=-3$ when $n$ leaves remainder $3$ on division by $6$ ; $x_n=3$ when it is divisible by $6$ ; otherwise $x_n=0$ Neither increasing nor decreasing	5
(5 marks)		5
Notes
Substitution gives the first six terms $0,0,-3,0,0,3$ . Both $2\cos(n\pi/3)$ and $(-1)^n$ repeat after $6$ steps, so $6$ is a period. Period $1$ fails since $x_2=0$ but $x_3=-3$ ; period $2$ fails since $x_1=0$ but $x_3=-3$ ; period $3$ fails since $x_3=-3$ but $x_6=3$ ; period $4$ fails since $x_2=0$ but $x_6=3$ ; and period $5$ fails since $x_1=0$ but $x_6=3$ . Therefore the least period is $6$ . Each repeated block shows that the value is $-3$ at positions leaving remainder $3$ on division by $6$ , $3$ at positions divisible by $6$ , and $0$ otherwise. Since, for example, $x_2>x_3$ but $x_3<x_4$ , the sequence is neither increasing nor decreasing.

4.3 · Understand and use sigma notation for sums of series.

Tier 1 · Easy

Mark scheme for 4.3 Tier 1 · Easy
Question	Scheme	Marks
1	$24$	2
(2 marks)		2
Notes
Substitute $r=1,2,3,4$ : the sum is $3+5+7+9=24$ .
2	$\displaystyle\sum_{r=1}^{13}(5r+6)$	2
(2 marks)		2
Notes
The term $5r+6$ gives $11$ when $r=1$ . Solving $5r+6=71$ gives $r=13$ , so the series is $\sum_{r=1}^{13}(5r+6)$ .

Tier 2 · Standard

Mark scheme for 4.3 Tier 2 · Standard
Question	Scheme	Marks
1	$540$	3
(3 marks)		3
Notes
Evaluate the eight terms directly. For $r=1,2,\ldots,8$ , the values of $3r^2-2r$ are $1,8,21,40,65,96,133,176$ . Their sum is $1+8+21+40+65+96+133+176=540$ .
2	$\displaystyle\sum_{s=1}^{9}(1-2s)$ $-81$	4
(4 marks)		4
Notes
Let $s=r-3$ . Then $r=4$ gives $s=1$ , $r=12$ gives $s=9$ , and $7-2r=7-2(s+3)=1-2s$ . Hence the sum is $\sum_{s=1}^{9}(1-2s)=9-2(9\times10/2)=-81$ .
3	$65$	3
(3 marks)		3
Notes
By linearity, the required sum is $2\sum_{r=1}^{7}a_r-3\sum_{r=1}^{7}b_r+\sum_{r=1}^{7}1$ . This is $2(23)-3(-4)+7=65$ .

Tier 3 · Hard

Mark scheme for 4.3 Tier 3 · Hard
Question	Scheme	Marks
1	$n=10$	5
(5 marks)		5
Notes
By linearity, $\sum_{r=1}^{n}(4r-1)=4[n(n+1)/2]-n=2n^2+n$ . Hence $2n^2+n=210$ , so $2n^2+n-210=0$ . Factorising gives $(2n+21)(n-10)=0$ . Since $n$ is positive, $n=10$ .
2	$k=4$	5
(5 marks)		5
Notes
There are $k+1$ terms, and $\sum_{r=k}^{2k}r=(k+2k)(k+1)/2=3k(k+1)/2$ . Therefore the given sum is $9k(k+1)/2-(k+1)=(k+1)(9k-2)/2$ . Equating this to $85$ gives $9k^2+7k-172=0=(k-4)(9k+43)$ . Since $k$ is a positive integer, $k=4$ .
3	$a=4$ $b=3$	4
(4 marks)		4
Notes
By linearity, the left side is $a\,n(n+1)/2+bn=(a/2)n^2+(a/2+b)n$ . The right side is $2n^2+5n$ . Since the identity holds for every positive integer $n$ , corresponding coefficients are equal. Thus $a/2=2$ , so $a=4$ , and then $a/2+b=5$ gives $b=3$ .
4	$S_n=3n$ Odd-upper-limit sum $=-3n-2$ $n=14$	5
(5 marks)		5
Notes
For each $j=1,\ldots,n$ , the terms with indices $2j-1$ and $2j$ sum to $-(6j-4)+(6j-1)=3$ . Hence $S_n=3n$ . The next term, with index $2n+1$ , is $-[3(2n+1)-1]=-(6n+2)$ , so the sum through $2n+1$ is $3n-(6n+2)=-3n-2$ . Setting this equal to $-44$ gives $-3n-2=-44$ , and therefore $n=14$ .
5	$n=10$	5
(5 marks)		5
Notes
For $r=1,2,3$ , the terms are $5,3,1$ , with sum $9$ . None of $n=1,2,3$ gives $58$ , so take $n\geq4$ . Then $\|2r-7\|=2r-7$ for $r\geq4$ , and $\sum_{r=4}^{n}(2r-7)=n(n+1)-12-7(n-3)=(n-3)^2$ . The equation is therefore $9+(n-3)^2=58$ , so $(n-3)^2=49$ . Since $n\geq4$ , $n-3=7$ and $n=10$ .

4.4 · Understand and work with arithmetic sequences and series, including the formulae for nth term and the sum to n terms.

Tier 1 · Easy

Mark scheme for 4.4 Tier 1 · Easy
Question	Scheme	Marks
1	$113$	2
(2 marks)		2
Notes
Here $a=11$ and $d=6$ . Thus $u_{18}=11+(18-1)6=11+102=113$ .
2	First term $5$ Fifteenth term $61$	3
(3 marks)		3
Notes
If the first term is $a$ , then $a+6(4)=29$ , so $a=5$ . The fifteenth term is $a+14d=5+14(4)=61$ .

Tier 2 · Standard

Mark scheme for 4.4 Tier 2 · Standard
Question	Scheme	Marks
1	First term $2$ Common difference $4$ $S_{20}=800$	5
(5 marks)		5
Notes
If the first term is $a$ and common difference is $d$ , then $a+4d=18$ and $a+11d=46$ . Subtracting gives $7d=28$ , so $d=4$ and then $a=2$ . Hence $S_{20}=\dfrac{20}{2}[2(2)+19(4)]=10(80)=800$ .
2	$u_n=6n-1$ The $28$ th term	4
(4 marks)		4
Notes
For $n\geq2$ , $u_n=S_n-S_{n-1}$ . Since $S_{n-1}=3(n-1)^2+2(n-1)=3n^2-4n+1$ , it follows that $u_n=6n-1$ ; this also gives $u_1=5=S_1$ . Solving $6n-1=167$ gives $n=28$ .
3	$14,21,28,35,42$ Sum $=140$	3
(3 marks)		3
Notes
Across the seven terms there are six equal steps, so the common difference is $(49-7)/6=7$ . The inserted terms are therefore $14,21,28,35,42$ , and their sum is $140$ .

Tier 3 · Hard

Mark scheme for 4.4 Tier 3 · Hard
Question	Scheme	Marks
1	$a=\dfrac{78}{5}$ $d=\dfrac{11}{5}$ $n=40$	5
(5 marks)		5
Notes
The term condition gives $a+7d=31$ . The sum condition gives $10(2a+19d)=730$ , so $2a+19d=73$ . Doubling the first equation and subtracting gives $5d=11$ , hence $d=11/5$ and $a=31-77/5=78/5$ . Therefore $u_n=(11n+67)/5$ . Solving $(11n+67)/5>100$ gives $n>433/11=39.36\ldots$ , so the least integer is $40$ .
2	$a=18$ $d=\dfrac32$ $S_{20}=645$	6
(6 marks)		6
Notes
The three lengths are $a$ , $a+4d$ and $a+8d$ , with the last as hypotenuse. Pythagoras gives $a^2+(a+4d)^2=(a+8d)^2$ , so $a^2-8ad-48d^2=0$ , or $(a-12d)(a+4d)=0$ . Positivity gives $a=12d$ . The perimeter condition is $3a+12d=72$ , hence $48d=72$ , so $d=3/2$ and $a=18$ . Therefore $S_{20}=10[2(18)+19(3/2)]=645$ .
3	$S_n$ is greatest when $n=8$ Greatest value $=128$ $4\leq n\leq12$ , $n$ an integer	5
(5 marks)		5
Notes
$S_n=\dfrac n2[60-4(n-1)]=2n(16-n)=128-2(n-8)^2$ . Thus the greatest value is $128$ at the positive integer $n=8$ . Also $S_n\geq96$ gives $2n(16-n)\geq96$ , or $(n-4)(n-12)\leq0$ . Hence the positive integers are $4\leq n\leq12$ .
4	$32,67,102,137$ Sum $=3633$	6
(6 marks)		6
Notes
Listing terms until the first match gives $7,12,17,22,27,32$ and $4,11,18,25,32$ , so the first common term is $32$ . The gap between successive common terms must be a multiple of both $5$ and $7$ . The positive multiples of $5$ below $35$ are $5,10,15,20,25,30$ , and none is a multiple of $7$ , while $35=5\times7$ . Hence the smallest possible positive gap is $35$ , and adding $35$ always produces another term of each sequence. The first four common terms are therefore $32,67,102,137$ . Those below $500$ are $32+35j$ for $j=0,1,\ldots,13$ , since the next term is $522$ . Their sum is $14[2(32)+13(35)]/2=3633$ .
5	$135$	5
(5 marks)		5
Notes
Let the hundreds digit be $a$ and the positive common difference be $d$ . The digits are $a,a+d,a+2d$ , so the integer is $100a+10(a+d)+(a+2d)=111a+12d$ . The digit sum is $3a+3d$ . The given condition gives $111a+12d=45a+45d$ , hence $d=2a$ . The units digit is then $a+2d=5a$ . Since $a$ is a positive digit and $5a\leq9$ , the only possibility is $a=1$ . Thus $d=2$ and the integer is $135$ , which indeed equals $15(1+3+5)$ .

4.5 · Understand and work with geometric sequences and series, including the formulae for the nth term and the sum of a finite geometric series; the sum to infinity of a convergent geometric series, including the use of |r| < 1; modulus notation.

Tier 1 · Easy

Mark scheme for 4.5 Tier 1 · Easy
Question	Scheme	Marks
1	$384$	2
(2 marks)		2
Notes
The first term is $a=3$ and the common ratio is $r=2$ . Hence $u_8=3\times2^7=384$ .
2	$k=18$	2
(2 marks)		2
Notes
For three consecutive geometric terms, the square of the middle term is the product of its neighbours. Thus $k^2=12\times27=324$ . Since $k$ is positive, $k=18$ .

Tier 2 · Standard

Mark scheme for 4.5 Tier 2 · Standard
Question	Scheme	Marks
1	First term $2$ Common ratio $3$ $S_5=242$	5
(5 marks)		5
Notes
With first term $a$ and ratio $r>0$ , $ar^2=18$ and $ar^5=486$ . Dividing gives $r^3=27$ , so $r=3$ . Then $a=18/3^2=2$ . Therefore $S_5=2(3^5-1)/(3-1)=2(242)/2=242$ .
2	$a=12$ $r=\dfrac25$	4
(4 marks)		4
Notes
From the sum to infinity, $a/(1-r)=20$ , so $a=20(1-r)$ . Also $a(1+r)=84/5$ . Substitution gives $20(1-r^2)=84/5$ , hence $r^2=4/25$ . The ratio is positive, so $r=2/5$ , and then $a=20(3/5)=12$ .
3	$\dfrac{26}{111}$	3
(3 marks)		3
Notes
The repeating blocks give the geometric series $234/1000+234/1000^2+\cdots$ , with ratio $r=1/1000$ . Since $\|r\|<1$ , its sum to infinity is $(234/1000)/(1-1/1000)=234/999=26/111$ .

Tier 3 · Hard

Mark scheme for 4.5 Tier 3 · Hard
Question	Scheme	Marks
1	$n=20$	5
(5 marks)		5
Notes
The finite sum is $S_n=800(1-0.9^n)/(1-0.9)=8000(1-0.9^n)$ . The condition $S_n>7000$ gives $0.9^n<0.125$ . Taking logarithms and accounting for $\ln(0.9)<0$ gives $n>\ln(0.125)/\ln(0.9)=19.73\ldots$ . Therefore the least integer is $n=20$ .
2	$(a,r)=(8,\dfrac12)$ or $(24,-\dfrac12)$ Both sums to infinity are $16$	6
(6 marks)		6
Notes
Since $S_4=S_2(1+r^2)$ , $15=12(1+r^2)$ , giving $r^2=1/4$ and hence $r=\pm1/2$ ; both satisfy $\|r\|<1$ . From $a(1+r)=12$ , $r=1/2$ gives $a=8$ , while $r=-1/2$ gives $a=24$ . The sums to infinity are respectively $8/(1-1/2)=16$ and $24/[1-(-1/2)]=16$ .
3	$\dfrac32<x<\dfrac{33}{16}$	5
(5 marks)		5
Notes
Convergence requires $\|x/3\|<1$ , so $-3<x<3$ . On this interval, $3-x>0$ and the sum is $S_\infty=(2x-1)/(1-x/3)=3(2x-1)/(3-x)$ . Solving $S_\infty>4$ gives $x>3/2$ , while $S_\infty<10$ gives $x<33/16$ . Both bounds lie inside the convergence interval, so $3/2<x<33/16$ .
4	First term $32$ ; common ratio $\dfrac12$ $S_6=63$ $S_\infty=64$	6
(6 marks)		6
Notes
If the first term is $a$ and ratio is $r$ , the odd-position sum is $a(1+r^2+r^4)=42$ . The even-position sum is $ar(1+r^2+r^4)=21$ , so division gives $r=1/2$ . Hence $a(1+1/4+1/16)=42$ , so $a(21/16)=42$ and $a=32$ . The first six terms have total $42+21=63$ . Since $\|r\|<1$ , the sum to infinity is $32/(1-1/2)=64$ .
5	$2-\sqrt3<k<1$ or $3<k<2+\sqrt3$ $S_\infty=\dfrac{5}{3-(k-2)^2}$ At $k=1$ or $k=3$ , the partial sums alternate between $5$ and $0$ ; at $k=2\pm\sqrt3$ , the partial sums are $5,10,15,\ldots$	6
(6 marks)		6
Notes
Since $q=(k-2)^2-2$ , convergence requires $-1<(k-2)^2-2<1$ . This is equivalent to $1<(k-2)^2<3$ , giving $2-\sqrt3<k<1$ or $3<k<2+\sqrt3$ . On this set, $S_\infty=5/(1-q)=5/[3-(k-2)^2]$ . At $k=1$ or $k=3$ , $q=-1$ , so the partial sums alternate between $5$ and $0$ and have no limit. At $k=2\pm\sqrt3$ , $q=1$ , so every term is $5$ and the partial sums grow as $5,10,15,\ldots$ . For every other value of $k$ , $\|q\|>1$ , so the terms do not tend to zero and the series diverges.

4.6 · Use sequences and series in modelling.

Tier 1 · Easy

Mark scheme for 4.6 Tier 1 · Easy
Question	Scheme	Marks
1	$\pounds 225$	2
(2 marks)		2
Notes
The monthly amounts form an arithmetic sequence with $a=25$ , $d=5$ and sixth term $25+5(5)=50$ . Therefore $S_6=6(25+50)/2=225$ .
2	$389$ litres	2
(2 marks)		2
Notes
The hourly multiplier is $0.92$ . After $3$ hours the model gives $500(0.92)^3=389.344$ litres, which is $389$ litres to $3$ significant figures.

Tier 2 · Standard

Mark scheme for 4.6 Tier 2 · Standard
Question	Scheme	Marks
1	$1000$ seats	3
(3 marks)		3
Notes
The row sizes form an arithmetic sequence with $a=16$ , $d=2$ and $n=25$ . The last row has $16+24(2)=64$ seats. Hence the total is $S_{25}=25(16+64)/2=1000$ seats.
2	$26$ complete frames $57$ tiles unused	4
(4 marks)		4
Notes
The total for $n$ frames is $S_n=\dfrac n2[2(18)+3(n-1)]=\dfrac{3n(n+11)}{2}$ . Now $S_{26}=3(26)(37)/2=1443$ , while $S_{27}=3(27)(38)/2=1539>1500$ . Therefore $26$ complete frames can be built, leaving $1500-1443=57$ tiles.
3	$T_3=47^\circ\mathrm{C}$ Steady temperature $=20^\circ\mathrm{C}$ For example, the model assumes an unchanged surrounding temperature and cooling multiplier.	4
(4 marks)		4
Notes
Iteration gives $T_1=68$ , $T_2=56$ and $T_3=47$ . Since $\|0.75\|<1$ , the recurrence converges to its fixed point. At a steady temperature $T$ , $T=0.75T+5$ , so $0.25T=5$ and $T=20^\circ\mathrm{C}$ . The fixed recurrence ignores changes in the surroundings or in the rate of cooling.

Tier 3 · Hard

Mark scheme for 4.6 Tier 3 · Hard
Question	Scheme	Marks
1	$B_n=5000(1.03)^n-400\dfrac{(1.03)^n-1}{0.03}$ $n=16$ For example, the model assumes the interest rate and withdrawal remain fixed and permits an unrealistic negative balance.	6
(6 marks)		6
Notes
After $n$ years, the initial balance has grown to $5000(1.03)^n$ . The withdrawals accumulate as $400[1+1.03+\cdots+1.03^{n-1}]=400[(1.03)^n-1]/0.03$ , giving the stated formula. Rearranging, $B_n=40000/3-(25000/3)(1.03)^n$ . A negative balance requires $(1.03)^n>1.6$ , so $n>\ln(1.6)/\ln(1.03)=15.90\ldots$ . The first integer is $n=16$ . In reality the account provider would not continue the same process into a negative balance, and the rate may change.
2	$49.8\%$ Limiting total from Sponsor B $=\pounds 4000$	6
(6 marks)		6
Notes
Sponsor A contributes $S_8=\dfrac82[2(200)+7(25)]=2300$ . Sponsor B contributes $S_8=400(1-0.9^8)/(1-0.9)=2278.13116$ . Therefore Sponsor B's percentage of the combined total is $100(2278.13116)/(2300+2278.13116)=49.7611\ldots\%$ , which is $49.8\%$ to $3$ significant figures. Since $\|0.9\|<1$ , Sponsor B's limiting total is $400/(1-0.9)=\pounds 4000$ .
3	Segments in stage $n$ : $6\times2^{n-1}$ Cable in stage $n$ : $54\left(\dfrac23\right)^{n-1}$ m First $5$ displays: $\dfrac{422}{3}$ m Limiting total $=162$ m For example, segments eventually become too short to manufacture.	6
(6 marks)		6
Notes
At each stage the number of segments doubles, so stage $n$ has $6\times2^{n-1}$ segments. Each segment then has length $9(1/3)^{n-1}$ metres, giving total stage length $54(2/3)^{n-1}$ metres. The first five displays use $54[1-(2/3)^5]/(1-2/3)=422/3$ metres. Since $\|2/3\|<1$ , the limiting total is $54/(1-2/3)=162$ metres. Indefinite continuation is unrealistic because the segments eventually become impractically short.
4	$a=288$ , $r=\dfrac12$ $n=9$ For example, the model assumes the same proportional decrease continues each day.	6
(6 marks)		6
Notes
The measurements give $ar^2=72$ and $ar^5=9$ . Dividing gives $r^3=1/8$ , so $r=1/2$ , and then $a=72/(1/2)^2=288$ . The condition is $288(1/2)^n<1$ , or $2^n>288$ . Since $2^8=256$ and $2^9=512$ , the first integer value is $n=9$ . The model ignores changes in removal rate, further pollution and any lower physical detection limit.
5	$D_n=\dfrac{100}{7}+\dfrac{40}{7}\left(\dfrac{3}{10}\right)^{n-1}$ Limiting daytime temperature $=\left(\dfrac{100}{7}\right)^\circ\mathrm{C}$ Limiting night-time temperature $=\left(\dfrac{74}{7}\right)^\circ\mathrm{C}$ For example, the model assumes the same heating and cooling rules every day.	7
(7 marks)		7
Notes
Substitution gives $D_{n+1}=0.5(0.6D_n+2)+9=0.3D_n+10$ . Its fixed value $L$ satisfies $L=0.3L+10$ , so $L=100/7$ . Therefore $D_{n+1}-100/7=(3/10)(D_n-100/7)$ . Since $D_1-100/7=40/7$ , it follows that $D_n=100/7+(40/7)(3/10)^{n-1}$ . As $\|3/10\|<1$ , $D_n$ tends to $100/7$ . Then $N_n=0.6D_n+2$ tends to $(3/5)(100/7)+2=74/7$ . Real greenhouse temperatures would also depend on weather, ventilation and changing heater performance.