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The binomial expansion

11 questions

LessonNot started
Find the term which is independent of xx in the expansion of (2x−3x2)6\left(2x - \dfrac{3}{x^{2}}\right)^{6}. (4 marks)
●●●●●Level 24 marksStart
Find the coefficient of x3x^3 in the expansion of (2−x)6(2 - x)^6.
●●●●●Level 32 marksStart
In the expansion of (3+kx)4(3 + kx)^4, the coefficient of x2x^2 is 216216. Find the possible values of the constant kk.
●●●●●Level 33 marksStart
Find the coefficient of x2x^2 in the expansion of (1+x)(2−x)4(1 + x)(2 - x)^4.
●●●●●Level 33 marksStart
A running club has 169 members who run mainly on roads and 91 members who run mainly on trails. A random sample of 18 members is selected to attend a meeting about the club's new training programme. The number of members in the sample who run mainly on trails is denoted by XX.
(a) State a suitable binomial model for XX.

(1 mark)

Use your model to answer the following.
(b) (i) Write down an expression for P(X=x)\mathrm{P}(X = x).

(ii) State, in set notation, the values of xx for which your expression is valid.

(iii) Explain the connection between the expression in part (b)(i) and the binomial expansion of (0.65+0.35)n(0.65 + 0.35)^n, for a specific value of nn which should be stated.

(4 marks)
(c) Find P(4⩽X⩽8)\mathrm{P}(4 \leqslant X \leqslant 8).

(2 marks)
(d) State one disadvantage of using a random sample in this context.

(1 mark)
●●●●●Level 48 marksStart
In this question you must show detailed reasoning.

The functions f and g are defined for all real values of xx by f(x)=x3f(x) = x^3 and g(x)=x2−4g(x) = x^2 - 4.
(a) Write down expressions for

(i) fg(x)fg(x),

(ii) gf(x)gf(x).

(2 marks)
(b) Hence find the values of xx for which fg(x)−gf(x)=−120fg(x) - gf(x) = -120.

(6 marks)
●●●●●Level 48 marksStart
(a) Find the first four terms in the expansion of (2−x)6(2 - x)^6 in ascending powers of xx.

(4 marks)
(b) Hence determine the coefficient of y3y^3 in the expansion of (2−y−3y2)6(2 - y - 3y^2)^6.

(4 marks)
●●●●●Level 48 marksStart
f(x)=(3+kx)8where k is a non-zero constant\text{f}(x) = (3 + kx)^8 \qquad \text{where } k \text{ is a non-zero constant}
(a) Find the first 4 terms, in ascending powers of xx, of the binomial expansion of f(x)\text{f}(x). Give each term in simplest form.

(4 marks)

Given that, in the binomial expansion of f(x)\text{f}(x), the constant term, the coefficient of xx and the coefficient of x2x^2 are the first 3 terms of an arithmetic progression,
(b) find, using algebra, the possible values of kk.

(Solutions relying entirely on calculator technology are not acceptable.)

(3 marks)
●●●●●Level 47 marksStart
The binomial expansion, in ascending powers of xx, of

(4+px)5(4 + px)^5

where pp is a constant, can be written in the form

A+Bx+Cx2+Dx3+…A + Bx + Cx^2 + Dx^3 + \ldots

where AA, BB, CC and DD are constants.
(a) Find the value of AA

(1 mark)

Given that
  • B=32DB = 32D
  • p<0p < 0
(b) find

(i) the value of pp

(ii) the value of DD

(6 marks)
●●●●●Level 47 marksStart
(a) Find, in terms of aa, the first 3 terms, in ascending powers of xx, of the binomial expansion of

(3+ax)5(3 + ax)^5

where aa is a positive constant. Give each term in simplest form.

(3 marks)

f(x)=(2x−1x)2(3+ax)5\text{f}(x) = \left(2x - \frac{1}{x}\right)^2(3 + ax)^5

Given that the constant term in the expansion of f(x)\text{f}(x) is 108108
(b) find the value of aa.

(4 marks)
●●●●●Level 47 marksStart
In this question you must show detailed reasoning.

The probability that a seed of a particular variety of sunflower fails to germinate is 0.24, independently of other seeds.
(a) A grower plants 400 of these seeds. The number of seeds that fail to germinate is denoted by the random variable YY. Determine the smallest value of aa such that P(Y>a)<0.05\mathrm{P}(Y > a) < 0.05.

(3 marks)

In the expansion of (0.24+0.76)40(0.24 + 0.76)^{40}, the terms involving 0.24r0.24^r and 0.24r+10.24^{r+1} are denoted by TrT_r and Tr+1T_{r+1} respectively.
(b) Show that Tr+1Tr=6(40−r)19(r+1)\dfrac{T_{r+1}}{T_r} = \dfrac{6(40 - r)}{19(r + 1)}.

(3 marks)

The number of seeds that fail to germinate in a packet of 40 of these seeds is modelled by the random variable XX.
(c) (i) Determine the values of rr for which P(X=r)⩽P(X=r+1)\mathrm{P}(X = r) \leqslant \mathrm{P}(X = r + 1).

(ii) Hence determine the most likely number of seeds in a packet of 40 that fail to germinate.

(6 marks)
●●●●●Level 512 marksStart

Arithmetic sequences and series

38 questions

LessonNot started
The third term of an arithmetic sequence is 1313 and the seventh term is 2929.
(a) Find the first term and the common difference. (2 marks)
(b) Find the sum of the first 2020 terms. (2 marks)
●●●●●Level 24 marksStart
The third term of an arithmetic sequence is 1313 and the seventh term is 2929.

Find the sum of the first 2020 terms of the sequence. (4 marks)
●●●●●Level 24 marksStart
(a) Show that

∑r=1n(6r+2)=3n2+5n\displaystyle\sum_{r=1}^{n} (6r + 2) = 3n^2 + 5n. (2 marks)
(b) Hence determine the value of nn for which

∑r=1n(6r+2)=100\displaystyle\sum_{r=1}^{n} (6r + 2) = 100. (3 marks)
●●●●●Level 25 marksStart
The sequence 3,7,11,15,…3, 7, 11, 15, \dots is arithmetic.
(a) Find an expression for the nnth term.
(b) Hence find the 5050th term.
●●●●●Level 23 marksStart
An arithmetic sequence has first term 88 and common difference 55.
(a) Find an expression for the nnth term. (2 marks)
(b) Find the 2020th term. (1 mark)
●●●●●Level 23 marksStart
Find the sum of the first 3030 terms of the arithmetic series

2+9+16+23+⋯2 + 9 + 16 + 23 + \cdots (3 marks)
●●●●●Level 23 marksStart
The nnth term of an arithmetic sequence is un=4n−1u_{n} = 4n - 1.

Find the value of nn for which un=199u_{n} = 199. (3 marks)
●●●●●Level 23 marksStart
The seventh term of an arithmetic progression is 1111. The sum of the first 1616 terms of the progression is 104104.

Determine the first term and the common difference.

(4 marks)
●●●●●Level 34 marksStart
An arithmetic progression has first term 1818 and common difference dd, where d≠0d \neq 0. The 1st, 4th and 6th terms of the arithmetic progression form, in that order, the first three terms of a geometric progression.

By determining dd, show that the arithmetic progression is a decreasing sequence.

(5 marks)
●●●●●Level 35 marksStart
(a) Show that

∑r=1n(3r+2)=n(3n+7)2\displaystyle\sum_{r=1}^{n} (3r + 2) = \frac{n(3n + 7)}{2}. (3 marks)
(b) Given that ∑r=1n(3r+2)=185\displaystyle\sum_{r=1}^{n} (3r + 2) = 185, find the value of nn, explaining why the other root of your quadratic must be rejected. (5 marks)
●●●●●Level 38 marksStart
An arithmetic sequence has first term aa and common difference dd. The third term is k+7k + 7 and the seventh term is 6k+46k + 4, where kk is a constant.

The sum of the first ten terms of the sequence is 175175.

Determine the values of kk, aa and dd. (6 marks)
●●●●●Level 36 marksStart
(a) Show that

∑r=1n(4r−1)=n(2n+1)\sum_{r=1}^{n} (4r - 1) = n(2n + 1)

(3 marks)
(b) Hence find the value of

∑r=1130(4r−1)\sum_{r=11}^{30} (4r - 1)

(3 marks)
●●●●●Level 36 marksStart
Find the sum of the first 2020 terms of the arithmetic series 3+7+11+…3 + 7 + 11 + \dots
●●●●●Level 32 marksStart
An arithmetic series has first term 55 and common difference 33. The sum of the first nn terms is 440440. Find nn.
●●●●●Level 34 marksStart
The 33rd term of an arithmetic sequence is 1313 and the 88th term is 3838.
(a) Find the first term and the common difference.
(b) Find the sum of the first 1010 terms.
●●●●●Level 34 marksStart
An arithmetic series has fourth term 1717, and the sum of its first six terms is 8787.

Find the first term and the common difference. (5 marks)
●●●●●Level 35 marksStart
An arithmetic series has first term 55 and common difference 44. The sum of the first nn terms is 275275.

Find nn. (5 marks)
●●●●●Level 35 marksStart
A graduate starts a job on a salary of £24,000\pounds 24{,}000, with a guaranteed rise of £1,500\pounds 1{,}500 each year.
(a) Find the salary in the 1010th year. (2 marks)
(b) Find the total amount earned over the first 1010 years. (3 marks)
●●●●●Level 35 marksStart
A fruit grower plants a new orchard. In its first year the orchard yields 18501850 kg of apples. The grower assumes that the yield will go up by the same amount every year. In the second year the orchard yields 19401940 kg of apples.
(a) Find the yield of the orchard in the twelfth year.

(2 marks)
(b) Find the number of complete years needed for the total yield of the orchard to first exceed 50 00050\,000 kg.

(4 marks)
(c) Comment on how realistic this model may be in the long term.

(1 mark)
●●●●●Level 47 marksStart
Nadia pays money into a savings account once a month. She pays in £a\pounds a in the first month, and each month after that she pays in £d\pounds d more than in the month before. In the fifth month she pays in £78\pounds 78. The total amount that she pays in over the first 1212 months is £1062\pounds 1062.
(a) Determine the values of aa and dd.

(4 marks)
(b) Find the number of complete months needed for the total amount that Nadia has paid in to first exceed £5000\pounds 5000.

(3 marks)
(c) Give a reason why this model may not be appropriate in the long term.

(1 mark)
●●●●●Level 48 marksStart
The first three terms of an arithmetic progression are the positive integers xx, yy and zz, in that order. The common difference of this progression is −3-3.

Also, xx, 20y\dfrac{20}{y} and zz are the first, second and third terms, respectively, of a geometric progression.
(a) Show that yy satisfies the equation y4−9y2−400=0y^4 - 9y^2 - 400 = 0.

(4 marks)
(b) Hence determine the sum to infinity of the geometric progression.

(4 marks)
●●●●●Level 48 marksStart
The first, second and fourth terms of an arithmetic progression are u1u_1, u2u_2 and u4u_4 respectively, where u1=3sin⁡θ,u2=5cos⁡θ,u4=−21sin⁡θu_1 = 3\sin\theta, \qquad u_2 = 5\cos\theta, \qquad u_4 = -21\sin\theta and 12π<θ<π\tfrac12\pi < \theta < \pi.
(a) Determine the exact value of θ\theta.

(3 marks)
(b) Hence determine the value of ∑r=120ur\displaystyle\sum_{r=1}^{20} u_r.

(3 marks)
●●●●●Level 46 marksStart
Two arithmetic progressions, PP and QQ, each have 6060 terms, denoted by pip_i and qiq_i respectively, where i=1,2,3,…,60i = 1, 2, 3, \ldots, 60.

The common difference of PP is dd, where dd is a positive integer. The progressions are related as follows.
  • p1=q60p_1 = q_{60}
  • q1=p60q_1 = p_{60}
(a) You are given that there is at least one value of ii for which qi=pi+24q_i = p_i + 24.

Show that, in this case, i=612−12di = \dfrac{61}{2} - \dfrac{12}{d}.

(6 marks)
(b) Hence show that it is impossible for the equation qi=pi+24q_i = p_i + 24 to hold unless dd takes certain values, which should be stated.

(2 marks)
●●●●●Level 48 marksStart
A small business makes handmade candles.
  • In year 1, the business made 15001500 candles.
  • In year 10, the business is predicted to make 54005400 candles.
Model A assumes that the number of candles made will increase by the same amount each year.
(a) Using model A, find the number of candles made in year 5. Give your answer to the nearest 10 candles.

(3 marks)

Model B assumes that the number of candles made will increase by the same percentage each year.
(b) Using model B, find the number of candles made in year 3. Give your answer to the nearest 10 candles.

(3 marks)
(c) Calculate, according to the two models, the difference between the total numbers of candles predicted to be made from year 1 to year 10 inclusive. Give your answer to the nearest 10 candles.

(3 marks)
●●●●●Level 49 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) A geometric series begins

45+30+20+…45 + 30 + 20 + \ldots

(i) Find the sum to infinity of this series.

(ii) Given that the sum to infinity of this series exceeds the sum of the first nn terms by less than 0.050.05, use algebra to find the smallest possible value of nn.

(5 marks)
(b) An arithmetic series begins

196+190.5+185+…196 + 190.5 + 185 + \ldots

Given that the sum of the first nn terms of this series is SnS_n, find the greatest possible value of SnS_n

(4 marks)
●●●●●Level 49 marksStart
f(x)=(3+kx)8where k is a non-zero constant\text{f}(x) = (3 + kx)^8 \qquad \text{where } k \text{ is a non-zero constant}
(a) Find the first 4 terms, in ascending powers of xx, of the binomial expansion of f(x)\text{f}(x). Give each term in simplest form.

(4 marks)

Given that, in the binomial expansion of f(x)\text{f}(x), the constant term, the coefficient of xx and the coefficient of x2x^2 are the first 3 terms of an arithmetic progression,
(b) find, using algebra, the possible values of kk.

(Solutions relying entirely on calculator technology are not acceptable.)

(3 marks)
●●●●●Level 47 marksStart
An arithmetic sequence has first term aa and non-zero common difference dd. The 22nd, 55th and 1111th terms of the arithmetic sequence form the first three terms of a geometric sequence.
(a) Show that the common ratio of the geometric sequence is 22. (5 marks)
(b) Given that a=4a = 4, find the sum of the first 2020 terms of the arithmetic sequence. (2 marks)
●●●●●Level 47 marksStart
The first three terms of a geometric progression are 4k4k, k+3k+3 and kk, where kk is a non-zero constant.
(a) Show that k2−2k−3=0k^{2} - 2k - 3 = 0 and hence find the two possible values of kk. (4 marks)
(b) It is given that the common ratio of the progression is positive. Find the sum to infinity. (2 marks)
(c) An arithmetic progression has first term equal to the first term of the geometric progression in part (b), and common difference 55. Find the least number of terms whose sum exceeds 500500. (3 marks)
●●●●●Level 49 marksStart
The first, second and fifth terms of an arithmetic sequence, with first term aa and common difference dd, where d≠0d \neq 0, are also the first three terms of a geometric sequence.

The sum of the first 2020 terms of the arithmetic sequence is 800800.

Find the smallest value of nn for which the nnth term of the geometric sequence is greater than one million. (6 marks)
●●●●●Level 46 marksStart
A charity receives a donation at the end of each week. The donation at the end of week 11 is £120120, and each week the donation is £4545 more than the donation the week before.
(a) Find the donation received at the end of week 3030. (2 marks)
(b) The charity's target is a total of £50 00050\,000. Find the number of the week at the end of which the running total first exceeds £50 00050\,000. (3 marks)
●●●●●Level 45 marksStart
A geometric sequence u1,u2,u3,…u_1, u_2, u_3, \ldots has first term u1=2u_1 = 2 and common ratio 44.
(a) Show that ∑r=1nlog⁡2ur=n2\displaystyle\sum_{r=1}^{n} \log_2 u_r = n^2. (3 marks)
(b) Hence determine the smallest value of nn for which

u1×u2×u3×⋯×un>1030u_1 \times u_2 \times u_3 \times \cdots \times u_n > 10^{30}. (4 marks)
●●●●●Level 47 marksStart
The sum of the first nn terms of a series is Sn=3n2+2nS_{n} = 3n^{2} + 2n.
(a) Find the first term. (1 mark)
(b) Show that the series is arithmetic, and find an expression for the nnth term and the common difference. (4 marks)
●●●●●Level 45 marksStart
Priya is training for a long-distance race.

In week 11 she runs a total of 1212 km. Each week she runs 22 km further than she ran the week before, so that the weekly distances form an arithmetic sequence.
(a) Show that Priya runs 3030 km in week 1010.

(1 mark)

Given that after nn weeks Priya has run a total of exactly 350350 km,
(b) show that

n2+11n−350=0n^2 + 11n - 350 = 0

(2 marks)
(c) Hence find the value of nn.

(2 marks)
●●●●●Level 45 marksStart
Two new podcasts, AA and BB, are launched on the same day. Each podcast is downloaded 500500 times in Week 1. The number of downloads of podcast AA will increase by 6060 for every subsequent week. The number of downloads of podcast BB will increase by 7%7\% for every subsequent week.
(a) Verify that in Week 10 podcast AA is downloaded 121121 more times than podcast BB, correct to the nearest whole number.

(3 marks)

Let Week XX be the first week in which podcast BB is downloaded more times than podcast AA.
(b) Show that XX satisfies the inequality X>1+log⁡1.07(0.12X+0.88)X > 1 + \log_{1.07}(0.12X + 0.88).

(3 marks)
(c) Use the iterative formula xn+1=1+log⁡1.07(0.12xn+0.88)x_{n+1} = 1 + \log_{1.07}(0.12x_n + 0.88) with x1=20x_1 = 20 to find the value of XX.

You should show the result of each iteration.

(3 marks)
(d) (i) Give a reason why the model for podcast AA may not be realistic.

(ii) Give a different reason why the model for podcast BB may not be realistic.

(2 marks)
●●●●●Level 511 marksStart
Anna saves money each week. She saves £15\pounds 15 in the first week, and increases the amount by £4\pounds 4 each week.
(a) Find how much she saves in week 2020. (2 marks)
(b) Find the total she has saved after 2020 weeks. (2 marks)
(c) Find the smallest number of weeks for her total savings to exceed £2000\pounds 2000. (4 marks)
●●●●●Level 58 marksStart
An arithmetic sequence has third term 1313 and eighth term 3333.
(a) Find the first term aa and common difference dd. (3 marks)
(b) Show that the sum of the first nn terms is Sn=2n2+3nS_n = 2n^2 + 3n. (3 marks)
●●●●●Level 56 marksStart
The sum of the first nn terms of a sequence is Sn=2n2+3nS_n = 2n^2 + 3n.
(a) Find an expression for the nnth term unu_n. (4 marks)
(b) Hence state, with a reason, whether the sequence is arithmetic. (2 marks)
●●●●●Level 56 marksStart
(a) In an arithmetic series, the first term is aa and the common difference is dd.

Prove that the sum of the first nn terms of the series is given by

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}\left[2a + (n - 1)d\right]

(3 marks)

Dev borrows £30003000 from a relative. He repays the loan in monthly instalments.

Dev repays £5555 in month 11, £6565 in month 22, £7575 in month 33 and so on, so that the monthly repayments form an arithmetic sequence.
(b) Show that Dev repays £145145 in month 1010.

(1 mark)

Given that Dev takes exactly nn months to repay the loan completely,
(c) show that

n2+10n−600=0n^2 + 10n - 600 = 0

(2 marks)
(d) Solve the equation in part (c).

(2 marks)
(e) Hence state the number of months Dev takes to repay the loan, giving a brief reason for your answer.

(1 mark)
●●●●●Level 59 marksStart

Geometric sequences and series

38 questions

LessonNot started
The sequence 2,6,18,54,…2, 6, 18, 54, \dots is geometric.
(a) Find an expression for the nnth term.
(b) Hence find the 66th term.
●●●●●Level 23 marksStart
An arithmetic progression has first term 1818 and common difference dd, where d≠0d \neq 0. The 1st, 4th and 6th terms of the arithmetic progression form, in that order, the first three terms of a geometric progression.

By determining dd, show that the arithmetic progression is a decreasing sequence.

(5 marks)
●●●●●Level 35 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Given that, in a particular geometric series,
  • the sum of the first three terms is 3737
  • the sum to infinity is 6464
find, for this series,
(a) the common ratio,

(4 marks)
(b) the first term.

(2 marks)
●●●●●Level 36 marksStart
A geometric series has first term 55 and common ratio 1.21.2.
(a) Find the 1212th term of the series, giving your answer to 33 significant figures. (2 marks)
(b) Find the least value of nn for which the sum of the first nn terms exceeds 10001000. (3 marks)
●●●●●Level 35 marksStart
The second term of a geometric progression is 66 and its sum to infinity is 3232.
(a) Show that the common ratio rr satisfies

16r2−16r+3=016r^2 - 16r + 3 = 0. (3 marks)
(b) Determine the possible values of the first term of the progression. (3 marks)
●●●●●Level 36 marksStart
Find the sum to infinity of the geometric series 8+4+2+1+…8 + 4 + 2 + 1 + \dots
●●●●●Level 32 marksStart
A geometric series has first term 55 and sum to infinity 2020. Find the common ratio rr.
●●●●●Level 32 marksStart
The 22nd term of a geometric series is 66 and the 55th term is 4848.
(a) Find the common ratio and the first term.
(b) Find the sum of the first 88 terms.
●●●●●Level 34 marksStart
A geometric series has first term 1212 and common ratio 23\tfrac23.
(a) Find the fifth term, to 3 s.f. (2 marks)
(b) Find the sum to infinity. (2 marks)
(c) Find the sum of the first six terms, to 3 s.f. (3 marks)
●●●●●Level 37 marksStart
A car is worth £20000\pounds 20000 when new. Each year its value falls by 15%15\% of its value at the start of that year.
(a) Show that its value after nn years is 20000(0.85)n20000(0.85)^n. (2 marks)
(b) Find its value after 55 years, to the nearest pound. (2 marks)
(c) Find the first year at the end of which the car is worth less than £8000\pounds 8000. (4 marks)
●●●●●Level 38 marksStart
A convergent geometric series has first term 2020 and sum to infinity 8080.
(a) Find the common ratio. (3 marks)
(b) Find the fourth term. (2 marks)
●●●●●Level 35 marksStart
A geometric series has first term 55 and common ratio (x−1)(x - 1).
(a) State the range of values of xx for which the sum to infinity exists. (2 marks)
(b) Given that the sum to infinity is 88, find the value of xx. (4 marks)
●●●●●Level 36 marksStart
A geometric series has first term 1212 and sum to infinity 1818.
(a) Find the common ratio. (3 marks)
(b) Find the sum of the first five terms, giving your answer correct to 3 significant figures. (3 marks)
●●●●●Level 36 marksStart
Nadia and Owen are playing a game.
  • The game consists of four rounds.
  • In each round, P(Nadia wins the round)=0.75\mathrm{P}(\text{Nadia wins the round}) = 0.75 and P(Owen wins the round)=0.25\mathrm{P}(\text{Owen wins the round}) = 0.25.
  • The outcome of each round is independent of the outcome of every other round.
  • If one player wins at least three of the four rounds, then they win the game, otherwise neither player wins.
(a) Determine the probability of the following events.

(i) Nadia wins the game.

(ii) Neither player wins the game.

(5 marks)

Nadia and Owen now decide to play a match which consists of a series of games.
  • In each game, if a player wins the game then they win the match.
  • If neither player wins the game then the players play another game.
(b) Determine the probability that Owen wins the match.

(3 marks)
●●●●●Level 48 marksStart
A sculptor builds a tower from wooden blocks, placing each block on top of the one before. The bottom block is 88 cm tall, and each block above it is 9%9\% shorter than the block directly beneath it.

The finished tower is 6060 cm tall, correct to the nearest centimetre.

Determine the number of blocks in the tower.

(5 marks)
●●●●●Level 45 marksStart
Imogen is learning to build bicycle wheels. Her first wheel takes her 8484 minutes to build. Her second wheel takes her 75.675.6 minutes to build. Imogen believes that the times she takes can be modelled by a geometric progression.
(a) Imogen sets herself a target of building a wheel in less than 3030 minutes. Show that this model predicts that Imogen will achieve her target on her eleventh wheel.

(4 marks)
(b) After some weeks, Imogen has spent a total of 603603 minutes, to the nearest minute, building wheels. Use this model to find how many wheels she has built.

(3 marks)
(c) Give two reasons why this model may not be appropriate when predicting the time Imogen takes to build a wheel.

(2 marks)
●●●●●Level 49 marksStart
In this question you must show detailed reasoning.

The first three terms of a convergent geometric progression are 3x+53x + 5, x+5x + 5 and 2x−52x - 5 respectively.

Determine the sum to infinity of this geometric progression.

(8 marks)
●●●●●Level 48 marksStart
In this question you must show detailed reasoning.

A sequence u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by un=18×0.75nu_n = 18 \times 0.75^n.

Use an algebraic method to find the smallest value of NN such that ∑n=1∞un−∑n=1Nun<10−3.\sum_{n=1}^{\infty} u_n - \sum_{n=1}^{N} u_n < 10^{-3}.

(8 marks)
●●●●●Level 48 marksStart
The first three terms of an arithmetic progression are the positive integers xx, yy and zz, in that order. The common difference of this progression is −3-3.

Also, xx, 20y\dfrac{20}{y} and zz are the first, second and third terms, respectively, of a geometric progression.
(a) Show that yy satisfies the equation y4−9y2−400=0y^4 - 9y^2 - 400 = 0.

(4 marks)
(b) Hence determine the sum to infinity of the geometric progression.

(4 marks)
●●●●●Level 48 marksStart
In this question you should assume that −1<x<1-1 < x < 1.
(a) (i) Find the first four terms in the expansion of (1+x)−2(1 + x)^{-2} in ascending powers of xx.

(ii) Write down the term in xnx^n in this expansion, where nn is a positive integer.

(3 marks)
(b) Write down the sum to infinity of the series 1−x+x2−x3+…1 - x + x^2 - x^3 + \ldots

(1 mark)
(c) Hence or otherwise find and simplify an expression for 2−5x+8x2−11x3+…2 - 5x + 8x^2 - 11x^3 + \ldots in the form a+bx(1+x)2\dfrac{a + bx}{(1 + x)^2}, where aa and bb are constants to be determined.

(3 marks)
●●●●●Level 47 marksStart
A small business makes handmade candles.
  • In year 1, the business made 15001500 candles.
  • In year 10, the business is predicted to make 54005400 candles.
Model A assumes that the number of candles made will increase by the same amount each year.
(a) Using model A, find the number of candles made in year 5. Give your answer to the nearest 10 candles.

(3 marks)

Model B assumes that the number of candles made will increase by the same percentage each year.
(b) Using model B, find the number of candles made in year 3. Give your answer to the nearest 10 candles.

(3 marks)
(c) Calculate, according to the two models, the difference between the total numbers of candles predicted to be made from year 1 to year 10 inclusive. Give your answer to the nearest 10 candles.

(3 marks)
●●●●●Level 49 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

A volunteer group is planting saplings as part of a rewilding project.

The numbers of saplings planted each month, in hundreds, for the first three months of the project were

k−4k3k−35k - 4 \qquad\qquad k \qquad\qquad 3k - 35

where kk is a constant.

Given that the numbers of saplings planted each month are modelled as a geometric series,
(a) show that 2k2−47k+140=02k^{2} - 47k + 140 = 0

(2 marks)
(b) predict the number of saplings planted in the 4th month.

(4 marks)

The total number of saplings planted is predicted to exceed 30 00030\,000 for the first time in the NNth month.
(c) Calculate the value of NN according to the model.

(3 marks)
●●●●●Level 49 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) A geometric series begins

45+30+20+…45 + 30 + 20 + \ldots

(i) Find the sum to infinity of this series.

(ii) Given that the sum to infinity of this series exceeds the sum of the first nn terms by less than 0.050.05, use algebra to find the smallest possible value of nn.

(5 marks)
(b) An arithmetic series begins

196+190.5+185+…196 + 190.5 + 185 + \ldots

Given that the sum of the first nn terms of this series is SnS_n, find the greatest possible value of SnS_n

(4 marks)
●●●●●Level 49 marksStart
The first three terms of a geometric sequence are

2k−54−kk−62k - 5 \qquad\qquad 4 - k \qquad\qquad k - 6

where kk is a constant.
(a) Show that kk satisfies the equation

k2−9k+14=0k^2 - 9k + 14 = 0

(2 marks)

Given that the sequence converges,
(b) (i) find the value of kk, giving a reason for your answer,

(ii) find the value of S∞S_\infty

(5 marks)
●●●●●Level 47 marksStart
The first three terms of a geometric series are

k+23k3k + 2\sqrt{3} \qquad\qquad k \qquad\qquad \sqrt{3}

where kk is a constant.

Given that the series converges,
(a) use algebra, making your reasoning clear, to prove that k=23k = 2\sqrt{3}

(3 marks)
(b) Find the sum to infinity of the series, giving your answer in the form a3a\sqrt{3}, where aa is an integer.

(2 marks)
(c) Find the smallest value of nn for which the sum of the first nn terms of the series exceeds 99%99\% of the sum to infinity.

(2 marks)
●●●●●Level 47 marksStart
The first three terms of a geometric series are

cos⁡θ1+sin⁡θ4cos⁡θ\cos\theta \qquad\qquad 1 + \sin\theta \qquad\qquad 4\cos\theta
(a) Show that

5sin⁡2θ+2sin⁡θ−3=05\sin^2\theta + 2\sin\theta - 3 = 0

(3 marks)

Given that 0<θ<π20 < \theta < \dfrac{\pi}{2},
(b) find the value of θ\theta, giving your answer to 3 decimal places.

(2 marks)
(c) Find the sum of the first 1010 terms of the series.

(3 marks)
●●●●●Level 48 marksStart
An arithmetic sequence has first term aa and non-zero common difference dd. The 22nd, 55th and 1111th terms of the arithmetic sequence form the first three terms of a geometric sequence.
(a) Show that the common ratio of the geometric sequence is 22. (5 marks)
(b) Given that a=4a = 4, find the sum of the first 2020 terms of the arithmetic sequence. (2 marks)
●●●●●Level 47 marksStart
A ball is dropped from a height of 22 m onto horizontal ground. After each bounce the ball rises to 65%65\% of the height it reached before that bounce.
(a) Find the height the ball reaches after the 55th bounce, giving your answer to 33 significant figures. (2 marks)
(b) Find the total distance travelled by the ball from the moment it is dropped until the moment it hits the ground for the 66th time, giving your answer to 33 significant figures. (4 marks)
(c) Assuming the ball continues to bounce indefinitely, find the total distance it travels, giving your answer to 33 significant figures. (3 marks)
●●●●●Level 49 marksStart
A geometric series has first term 1212 and sum to infinity 1616.
(a) Show that the common ratio of the series is 14\dfrac{1}{4}. (3 marks)
(b) Find the smallest value of nn for which the sum to infinity differs from the sum of the first nn terms by less than 0.0010.001. (5 marks)
●●●●●Level 48 marksStart
The first three terms of a geometric progression are 4k4k, k+3k+3 and kk, where kk is a non-zero constant.
(a) Show that k2−2k−3=0k^{2} - 2k - 3 = 0 and hence find the two possible values of kk. (4 marks)
(b) It is given that the common ratio of the progression is positive. Find the sum to infinity. (2 marks)
(c) An arithmetic progression has first term equal to the first term of the geometric progression in part (b), and common difference 55. Find the least number of terms whose sum exceeds 500500. (3 marks)
●●●●●Level 49 marksStart
A company makes a profit of £150 000150\,000 in its first year of trading. In a model, the profit is assumed to grow by 4%4\% each year, so the annual profits form a geometric sequence.
(a) Find the profit the model predicts for year 1010. Give your answer to the nearest pound. (2 marks)
(b) Find the total profit the model predicts for the first 1515 years of trading. Give your answer to the nearest pound. (3 marks)
(c) Find the first year in which the model predicts an annual profit of more than £250 000250\,000. Show your working clearly. (3 marks)
●●●●●Level 48 marksStart
A geometric series has common ratio rr, where ∣r∣<1|r| < 1. The sum of the first two terms of the series is 99, and the sum to infinity of the series is 2525.
(a) Find the two possible values of rr. (4 marks)
(b) For the series with the smaller first term, find the least value of nn for which the sum of the first nn terms exceeds 24.524.5. (2 marks)
●●●●●Level 46 marksStart
The first, second and fifth terms of an arithmetic sequence, with first term aa and common difference dd, where d≠0d \neq 0, are also the first three terms of a geometric sequence.

The sum of the first 2020 terms of the arithmetic sequence is 800800.

Find the smallest value of nn for which the nnth term of the geometric sequence is greater than one million. (6 marks)
●●●●●Level 46 marksStart
A geometric sequence u1,u2,u3,…u_1, u_2, u_3, \ldots has first term u1=2u_1 = 2 and common ratio 44.
(a) Show that ∑r=1nlog⁡2ur=n2\displaystyle\sum_{r=1}^{n} \log_2 u_r = n^2. (3 marks)
(b) Hence determine the smallest value of nn for which

u1×u2×u3×⋯×un>1030u_1 \times u_2 \times u_3 \times \cdots \times u_n > 10^{30}. (4 marks)
●●●●●Level 47 marksStart
A town council believes that the use of its new cycle-hire scheme can be modelled by H=abtH = ab^t, where aa and bb are both positive constants, HH is the number of hires in a week and tt is the number of complete weeks since the scheme opened.

The numbers of hires in the first six weeks are recorded, and the values of log⁡10H\log_{10} H are plotted against tt in the graph below.1234562.32.42.52.62.72.82.9tlog₁₀ H(a) Explain why the graph suggests that the given model is appropriate.

(3 marks)

The council believes that a=250a = 250 and b=1.2b = 1.2 are good estimates for the parameters in the model.
(b) Show that the graph supports these estimates for the parameters.

(2 marks)
(c) Use the model H=250×1.2tH = 250 \times 1.2^t to predict the number of hires in the tenth week after the scheme opened.

(2 marks)
(d) (i) Use the model H=250×1.2tH = 250 \times 1.2^t to predict the number of weeks after the scheme opened when the total number of hires since opening will first exceed 40 000.

(ii) Comment on how reliable this prediction may be.

(5 marks)
●●●●●Level 512 marksStart
Two new podcasts, AA and BB, are launched on the same day. Each podcast is downloaded 500500 times in Week 1. The number of downloads of podcast AA will increase by 6060 for every subsequent week. The number of downloads of podcast BB will increase by 7%7\% for every subsequent week.
(a) Verify that in Week 10 podcast AA is downloaded 121121 more times than podcast BB, correct to the nearest whole number.

(3 marks)

Let Week XX be the first week in which podcast BB is downloaded more times than podcast AA.
(b) Show that XX satisfies the inequality X>1+log⁡1.07(0.12X+0.88)X > 1 + \log_{1.07}(0.12X + 0.88).

(3 marks)
(c) Use the iterative formula xn+1=1+log⁡1.07(0.12xn+0.88)x_{n+1} = 1 + \log_{1.07}(0.12x_n + 0.88) with x1=20x_1 = 20 to find the value of XX.

You should show the result of each iteration.

(3 marks)
(d) (i) Give a reason why the model for podcast AA may not be realistic.

(ii) Give a different reason why the model for podcast BB may not be realistic.

(2 marks)
●●●●●Level 511 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

The numbers of electric cars and petrol cars registered in a town are being monitored.

When monitoring began, there were 15001500 electric cars registered in the town.

A model predicts that the number of electric cars registered in the town will increase by 16%16\% each year, so that the numbers of electric cars at the end of each year form a geometric sequence.
(a) Find, according to the model, the number of electric cars registered in the town 6 years after monitoring began. Give your answer to 3 significant figures.

(2 marks)

The number of petrol cars registered in the town is monitored over the same period of time.

Given that
  • 3 years after monitoring began there were 16 60016\,600 petrol cars registered in the town
  • 8 years after monitoring began there were 12 20012\,200 petrol cars registered in the town
  • the number of petrol cars registered in the town at the end of each year is modelled as a geometric sequence with equation N=abtN = ab^{t}, where NN is the number of petrol cars tt years after monitoring began and aa and bb are constants
(b) (i) show that b=0.94b = 0.94 to 2 significant figures,

(ii) find the value of aa, giving your answer to 2 significant figures.

(3 marks)

When t=Tt = T, the number of electric cars registered in the town is equal to the number of petrol cars registered in the town.
(c) Using b=0.94b = 0.94 and your value of aa, find, according to the models, the value of TT, giving your answer to one decimal place.

(3 marks)
(d) Find, according to the model for petrol cars, the smallest whole number of years after monitoring began at which there are fewer than 80008000 petrol cars registered in the town.

(3 marks)
●●●●●Level 511 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Given that the first three terms of a geometric series are

22cos⁡θ7+8sin⁡θ11tan⁡θ22\cos\theta \qquad\qquad 7 + 8\sin\theta \qquad\qquad 11\tan\theta
(a) show that

64sin⁡2θ−130sin⁡θ+49=064\sin^2\theta - 130\sin\theta + 49 = 0

(3 marks)

Given that θ\theta is an obtuse angle measured in radians,
(b) solve the equation in part (a) to find the exact value of θ\theta

(2 marks)
(c) Find the sum to infinity of the series, giving your answer in the form a+b3a + b\sqrt{3}, where aa and bb are rational constants to be found.

(5 marks)
●●●●●Level 510 marksStart

Sigma notation and recurrence relations

29 questions

LessonNot started
A sequence u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by

u1=3u_1 = 3

un+1=un+1un−1u_{n+1} = \frac{u_n + 1}{u_n - 1}
(a) (i) Show that this sequence is periodic.

(ii) State the order of this periodic sequence.

(2 marks)
(b) Find the value of ∑n=175un\displaystyle\sum_{n=1}^{75} u_n

(2 marks)
●●●●●Level 24 marksStart
A sequence u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by

u1=ku_1 = k

un+1=8−unu_{n+1} = 8 - u_n

where kk is a constant.
(a) Find, in terms of kk,

(i) u2u_2

(ii) u3u_3

(2 marks)

Given that

∑n=141un=173\sum_{n=1}^{41} u_n = 173
(b) find the value of kk.

(2 marks)
●●●●●Level 24 marksStart
(a) Show that

∑r=1n(6r+2)=3n2+5n\displaystyle\sum_{r=1}^{n} (6r + 2) = 3n^2 + 5n. (2 marks)
(b) Hence determine the value of nn for which

∑r=1n(6r+2)=100\displaystyle\sum_{r=1}^{n} (6r + 2) = 100. (3 marks)
●●●●●Level 25 marksStart
A sequence is defined by u1=5u_{1} = 5 and un+1=un+4u_{n+1} = u_{n} + 4.
(a) Write down u2u_{2}, u3u_{3} and u4u_{4}. (2 marks)
(b) Find the value of nn for which un=101u_{n} = 101. (2 marks)
●●●●●Level 24 marksStart
A sequence has terms u1,u2,u3,…u_1, u_2, u_3, \ldots defined by u1=5u_1 = 5 and un+1=un−3un+1u_{n+1} = \dfrac{u_n - 3}{u_n + 1} for n⩾1n \geqslant 1.
(a) Find the values of u2u_2, u3u_3 and u4u_4.

(2 marks)
(b) Describe the behaviour of the sequence.

(1 mark)
●●●●●Level 33 marksStart
(a) Show that

∑r=1n(3r+2)=n(3n+7)2\displaystyle\sum_{r=1}^{n} (3r + 2) = \frac{n(3n + 7)}{2}. (3 marks)
(b) Given that ∑r=1n(3r+2)=185\displaystyle\sum_{r=1}^{n} (3r + 2) = 185, find the value of nn, explaining why the other root of your quadratic must be rejected. (5 marks)
●●●●●Level 38 marksStart
A sequence is defined by

u1=ku_1 = k, un+1=5−un\quad u_{n+1} = 5 - u_n,

where kk is a constant.
(a) Show that u3=u1u_3 = u_1, and describe the behaviour of the sequence. (2 marks)
(b) Given that ∑r=1101ur=258\displaystyle\sum_{r=1}^{101} u_r = 258, find the value of kk. (4 marks)
●●●●●Level 36 marksStart
(a) Show that

∑r=1n(4r−1)=n(2n+1)\sum_{r=1}^{n} (4r - 1) = n(2n + 1)

(3 marks)
(b) Hence find the value of

∑r=1130(4r−1)\sum_{r=11}^{30} (4r - 1)

(3 marks)
●●●●●Level 36 marksStart
Evaluate ∑r=110(2r+1)\displaystyle\sum_{r=1}^{10} (2r + 1).
●●●●●Level 32 marksStart
A sequence is defined by un+1=2un−1u_{n+1} = 2u_n - 1 with u1=3u_1 = 3. Find u2u_2, u3u_3 and u4u_4.
●●●●●Level 32 marksStart
Evaluate ∑r=120(3r−2)\displaystyle\sum_{r=1}^{20} (3r - 2).
●●●●●Level 32 marksStart
(a) Evaluate ∑r=115(5r−3)\displaystyle\sum_{r=1}^{15} (5r - 3). (3 marks)
(b) Hence evaluate ∑r=615(5r−3)\displaystyle\sum_{r=6}^{15} (5r - 3). (2 marks)
●●●●●Level 35 marksStart
A sequence is defined by u1=2u_{1} = 2 and un+1=1unu_{n+1} = \dfrac{1}{u_{n}}.
(a) Write down u2u_{2} and u3u_{3}, and describe the behaviour of the sequence. (2 marks)
(b) Write down u50u_{50}. (1 mark)
(c) Find ∑n=150un\displaystyle\sum_{n=1}^{50} u_{n}. (2 marks)
●●●●●Level 35 marksStart
(a) Show that ∑r=1n(2r+3)=n(n+4)\displaystyle\sum_{r=1}^{n} (2r + 3) = n(n + 4). (3 marks)
(b) Hence find the value of nn for which ∑r=1n(2r+3)=140\displaystyle\sum_{r=1}^{n} (2r + 3) = 140. (2 marks)
●●●●●Level 35 marksStart
A sequence is defined by u1=10u_{1} = 10 and un+1=0.8un+6u_{n+1} = 0.8u_{n} + 6.
(a) Find u2u_{2} and u3u_{3}. (2 marks)
(b) The sequence converges to a limit LL. Find the value of LL. (2 marks)
(c) Describe the behaviour of the sequence. (1 mark)
●●●●●Level 35 marksStart
Evaluate ∑r=1030(4r+1)\displaystyle\sum_{r=10}^{30} (4r + 1). (4 marks)
●●●●●Level 34 marksStart
A lake contains 100100 tonnes of algae. Each week, 10%10\% of the algae dies off naturally, and 3030 tonnes of new algae grows. The mass after nn weeks is modelled by u1=100u_{1} = 100 and

un+1=0.9un+30u_{n+1} = 0.9u_{n} + 30
(a) Find the mass after each of the next two weeks. (2 marks)
(b) The mass approaches a limit LL. Find LL. (2 marks)
(c) Interpret LL in the context of the model. (1 mark)
●●●●●Level 35 marksStart
(a) Show that ∑r=1n(6r+2)=3n2+5n\displaystyle\sum_{r=1}^{n} (6r + 2) = 3n^{2} + 5n. (2 marks)
(b) Hence find the value of nn for which ∑r=1n(6r+2)=848\displaystyle\sum_{r=1}^{n} (6r + 2) = 848. (3 marks)
●●●●●Level 35 marksStart
A sequence has terms u1,u2,u3,…u_1, u_2, u_3, \ldots defined by u1=3u_1 = 3 and un+1=un+6un−1u_{n+1} = \dfrac{u_n + 6}{u_n - 1} for n⩾1n \geqslant 1.
(a) Find the values of u2u_2, u3u_3 and u4u_4.

(2 marks)
(b) Describe the behaviour of the sequence.

(1 mark)
(c) Given that ∑n=1kun=363\displaystyle\sum_{n=1}^{k} u_n = 363, determine the value of kk.

(3 marks)
●●●●●Level 46 marksStart
A sequence of numbers u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by

u1=5u_1 = 5

un+1=(−1)n+1un+ku_{n+1} = (-1)^{n+1}u_n + k

where kk is a constant.
(a) Show that u5=5u_5 = 5

(3 marks)

Given that

∑r=126ur=88\sum_{r=1}^{26} u_r = 88
(b) find the value of kk.

(2 marks)
(c) Hence find the value of

∑r=1103ur\sum_{r=1}^{103} u_r

(2 marks)
●●●●●Level 47 marksStart
A sequence u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by

un+1=aun−bu1=4u_{n+1} = au_n - b \qquad\qquad u_1 = 4

where aa and bb are constants.
(a) Find, in terms of aa and bb,

(i) u2u_2

(ii) u3u_3

(2 marks)

Given
  • ∑n=13un=36\displaystyle\sum_{n=1}^{3} u_n = 36
  • b=2a+5b = 2a + 5
(b) show that

2a2−5a−42=02a^{2} - 5a - 42 = 0

(3 marks)
(c) Hence find the smaller possible value of u4u_4

(3 marks)
●●●●●Level 48 marksStart
A sequence u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by

u1=9u_1 = 9

un+1=un+3cos⁡(nπ2)u_{n+1} = u_n + 3\cos\left(\frac{n\pi}{2}\right)
(a) Find the value of u2u_2, the value of u3u_3, the value of u4u_4 and the value of u5u_5

(2 marks)
(b) State the order of this periodic sequence.

(1 mark)
(c) Find the value of ∑n=162un\displaystyle\sum_{n=1}^{62} u_n

(3 marks)
●●●●●Level 46 marksStart
A sequence a1,a2,a3,…a_1, a_2, a_3, \ldots is defined by

a1=ka_1 = k, an+1=1−1an\qquad a_{n+1} = 1 - \dfrac{1}{a_n}, n≥1\quad n \ge 1,

where kk is a constant such that every term of the sequence is defined.
(a) Show that a4=ka_4 = k. (4 marks)
(b) Explain why kk cannot be equal to 11. (1 mark)
(c) Given that k=3k = 3, find the value of

∑r=1200(2ar−1)\displaystyle\sum_{r=1}^{200} \left(2a_r - 1\right),

giving your answer as an exact fraction. (3 marks)
●●●●●Level 48 marksStart
A sequence u1,u2,u3,…u_1, u_2, u_3, \ldots is defined by

u1=4,un+1=un−1un,n⩾1u_1 = 4, \qquad u_{n+1} = \dfrac{u_n - 1}{u_n}, \qquad n \geqslant 1
(a) Show that the sequence is periodic, and state its period. (2 marks)
(b) Find the exact value of ∑r=1200ur\displaystyle\sum_{r=1}^{200} u_r (3 marks)
●●●●●Level 45 marksStart
A geometric sequence u1,u2,u3,…u_1, u_2, u_3, \ldots has first term u1=2u_1 = 2 and common ratio 44.
(a) Show that ∑r=1nlog⁡2ur=n2\displaystyle\sum_{r=1}^{n} \log_2 u_r = n^2. (3 marks)
(b) Hence determine the smallest value of nn for which

u1×u2×u3×⋯×un>1030u_1 \times u_2 \times u_3 \times \cdots \times u_n > 10^{30}. (4 marks)
●●●●●Level 47 marksStart
A sequence is defined by u1=3u_1 = 3 and un+1=2un−1u_{n+1} = 2u_n - 1.
(a) Find u2u_2, u3u_3 and u4u_4. (3 marks)
(b) Evaluate ∑n=14un\displaystyle\sum_{n=1}^{4} u_n. (2 marks)
●●●●●Level 45 marksStart
A sequence is defined by u1=3u_{1} = 3 and un+1=kun+2u_{n+1} = ku_{n} + 2, where kk is a constant.
(a) Find expressions for u2u_{2} and u3u_{3} in terms of kk. (2 marks)
(b) Given that u3=10u_{3} = 10 and that kk is an integer, find the value of kk. (3 marks)
●●●●●Level 45 marksStart
A sequence is defined by u1=4u_{1} = 4 and un+1=2un−3un−1u_{n+1} = \dfrac{2u_{n} - 3}{u_{n} - 1}.
(a) Show that the sequence is periodic, and state its period. (2 marks)
(b) Write down u99u_{99}. (1 mark)
(c) Find ∑n=190un\displaystyle\sum_{n=1}^{90} u_{n}. (2 marks)
●●●●●Level 45 marksStart
A sequence is given by un+1=4−unu_{n+1} = 4 - u_n with u1=1u_1 = 1.
(a) Find u2u_2, u3u_3 and u4u_4.
(b) State the period of the sequence.
(c) Hence find ∑r=150ur\displaystyle\sum_{r=1}^{50} u_r.
●●●●●Level 55 marksStart

Binomial expansion for any index

35 questions

LessonNot started
The binomial expansion of (1+2x)−1(1 + 2x)^{-1} is valid for

Circle your answer.
  • A ∣x∣<2|x| < 2
  • B ∣x∣<1|x| < 1
  • C ∣x∣<12|x| < \dfrac{1}{2}
  • D ∣x∣≤12|x| \le \dfrac{1}{2}
●●●●●Level 11 markStart
The binomial expansion of 14−3x\dfrac{1}{\sqrt{4 - 3x}} in ascending powers of xx is valid for

Circle your answer.
  • A ∣x∣<34|x| < \dfrac{3}{4}
  • B ∣x∣<43|x| < \dfrac{4}{3}
  • C ∣x∣<4|x| < 4
  • D ∣x∣<14|x| < \dfrac{1}{4}
●●●●●Level 11 markStart
(a) Find the first three terms, in ascending powers of xx, of the binomial expansion of

9−8x\sqrt{9 - 8x}

writing each term in simplest form.

(3 marks)

A student uses this expansion with x=14x = \dfrac{1}{4} to find an approximation for 7\sqrt{7}

Using the answer to part (a) and without doing any calculations,
(b) state whether this approximation will be an overestimate or an underestimate of 7\sqrt{7}, giving a brief reason for your answer.

(1 mark)
●●●●●Level 24 marksStart
(a) Find the first three terms, in ascending powers of xx, of the binomial expansion of

(8−3x)13(8 - 3x)^{\frac{1}{3}}

writing each term in simplest form.

(3 marks)

A student attempts to use this expansion with x=−3x = -3 to find an approximation for 173\sqrt[3]{17}
(b) Give a reason why x=−3x = -3 should not be used in this expansion to find an approximation for 173\sqrt[3]{17}

(1 mark)
●●●●●Level 24 marksStart
(a) Find the first three terms, in ascending powers of xx, of the binomial expansion of

(8+3x)23(8+3x)^{\frac{2}{3}},

and state the range of values of xx for which the expansion is valid. (4 marks)
(b) Use your expansion to estimate 8.3238.3^{\frac{2}{3}}, giving your answer to 4 decimal places. (1 mark)
●●●●●Level 25 marksStart
f(x)=4−xf(x) = \sqrt{4 - x}.
(a) Find the binomial expansion of f(x)f(x) in ascending powers of xx, up to and including the term in x2x^2, and state the range of values of xx for which the expansion is valid. (3 marks)
(b) Use your expansion to find an approximate value for 3.9\sqrt{3.9}, giving your answer to 5 decimal places. (1 mark)
●●●●●Level 34 marksStart
(a) Find the first three terms, in ascending powers of xx, of the binomial expansion of

14−3x\dfrac{1}{\sqrt{4 - 3x}}

giving each coefficient in its simplest form. (3 marks)
(b) State the range of values of xx for which this expansion is valid. (1 mark)
●●●●●Level 34 marksStart
SECTION A: PURE MATHEMATICS
(a) Find the first three terms, in ascending powers of xx, of the binomial expansion of 1−4x\sqrt{1 - 4x}, giving each coefficient in its simplest form. (4 marks)
(b) State the range of values of xx for which this expansion is valid. (1 mark)
●●●●●Level 35 marksStart
Expand (1+x)−1(1 + x)^{-1} up to and including the term in x3x^3, and state the range of values of xx for which the expansion is valid.
●●●●●Level 33 marksStart
Expand (1+2x)−3(1 + 2x)^{-3} in ascending powers of xx, up to and including the term in x3x^{3}, and state the range of xx for which the expansion is valid. (4 marks)
●●●●●Level 34 marksStart
(a) Expand (9−3x)12(9 - 3x)^{\frac{1}{2}} in ascending powers of xx, up to and including the term in x2x^{2}, and state the range of validity. (4 marks)
(b) By substituting x=0.3x = 0.3, find an approximation to 8.1\sqrt{8.1}, giving your answer to 55 decimal places. (2 marks)
●●●●●Level 36 marksStart
(a) Expand (1−4x)12(1 - 4x)^{\frac{1}{2}} in ascending powers of xx, up to and including the term in x2x^{2}. (3 marks)
(b) Explain why the expansion cannot be used to estimate 1−4(0.3)\sqrt{1 - 4(0.3)}. (1 mark)
(c) Use the expansion with a suitable value of xx to estimate 0.92\sqrt{0.92}, giving your answer to 44 decimal places. (2 marks)
●●●●●Level 36 marksStart
Expand 12+x\dfrac{1}{2 + x} in ascending powers of xx, up to and including the term in x3x^{3}, and state the range of validity. (4 marks)
●●●●●Level 34 marksStart
(a) Find the first three terms in the expansion of (4−3x)−12(4 - 3x)^{-\frac{1}{2}} in ascending powers of xx.

(4 marks)
(b) State the range of values of xx for which the expansion in part (a) is valid.

(1 mark)
(c) Find the coefficient of x2x^2 in the expansion of 2+5x4−3x\dfrac{2 + 5x}{\sqrt{4 - 3x}}.

(4 marks)
●●●●●Level 49 marksStart
(a) Find the first three terms in the expansion of (4−16x)12(4 - 16x)^{\frac{1}{2}} in ascending powers of xx.

(4 marks)
(b) State the range of values of xx for which the expansion in part (a) is valid.

(1 mark)
(c) In the expansion of (1+ax)2(4−16x)12(1 + ax)^2(4 - 16x)^{\frac{1}{2}} the coefficient of x2x^2 is 66.

Determine the possible values of the constant aa.

(4 marks)
●●●●●Level 49 marksStart
(a) Find the first three terms in the expansion of (9+2x)−12(9 + 2x)^{-\frac{1}{2}} in ascending powers of xx.

(4 marks)
(b) State the set of values of xx for which this expansion is valid.

(1 mark)
(c) When 6+ax9+2x\dfrac{6 + ax}{\sqrt{9 + 2x}} is expanded in ascending powers of xx, the coefficient of xx is zero.

Determine the value of the constant aa.

(2 marks)
●●●●●Level 47 marksStart
The diagram shows triangle ABCABC in which angle BACBAC is 120°120° and the lengths of ABAB and ACAC are (5+h)(5 + h) cm and (3−h)(3 - h) cm respectively, where 0<h<30 < h < 3.(5 + h) cm(3 − h) cm120°ABC(a) Show that the length of BCBC is p\sqrt{p} cm where p=49+2h+h2p = 49 + 2h + h^2.

(2 marks)
(b) Hence show that, when hh is small, p≈7+mh+nh2\sqrt{p} \approx 7 + mh + nh^2, where mm and nn are rational numbers whose values are to be determined.

(4 marks)
●●●●●Level 46 marksStart
In this question you should assume that −1<x<1-1 < x < 1.
(a) (i) Find the first four terms in the expansion of (1+x)−2(1 + x)^{-2} in ascending powers of xx.

(ii) Write down the term in xnx^n in this expansion, where nn is a positive integer.

(3 marks)
(b) Write down the sum to infinity of the series 1−x+x2−x3+…1 - x + x^2 - x^3 + \ldots

(1 mark)
(c) Hence or otherwise find and simplify an expression for 2−5x+8x2−11x3+…2 - 5x + 8x^2 - 11x^3 + \ldots in the form a+bx(1+x)2\dfrac{a + bx}{(1 + x)^2}, where aa and bb are constants to be determined.

(3 marks)
●●●●●Level 47 marksStart
(a) Find the binomial expansion of (8−3x)13(8 - 3x)^{\frac13} in ascending powers of xx, up to and including the term in x2x^2, giving each coefficient in its simplest form. (4 marks)
(b) Use your expansion, with a suitable value of xx, to find an estimate for 7.73\sqrt[3]{7.7}, giving your answer to 44 decimal places. (2 marks)
●●●●●Level 46 marksStart
Expand 4−3x\sqrt{4 - 3x} in ascending powers of xx, up to and including the term in x2x^{2}, simplifying the coefficients.

State the set of values of xx for which the expansion is valid. (5 marks)
●●●●●Level 45 marksStart
Given that the binomial expansion, in ascending powers of xx, of

(8+Ax)13(8 + Ax)^{\frac{1}{3}}

is

2+Bx+Cx2+512x3+…2 + Bx + Cx^2 + \dfrac{5}{12}x^3 + \ldots
(a) find the values of the constants AA, BB and CC.

(5 marks)
(b) State the range of values of xx for which this expansion is valid.

(1 mark)

For the expansion,
(c) find the coefficient of the term in x4x^4

(2 marks)
●●●●●Level 48 marksStart
Find the first three terms, in ascending powers of xx, of the expansion of (1+2x)1/2(1 + 2x)^{1/2}, and state the range of validity.
●●●●●Level 43 marksStart
Expand (1−3x)−2(1 - 3x)^{-2} up to and including the term in x2x^2, stating the range of validity.
●●●●●Level 43 marksStart
(a) Expand (1−2x)−2(1 - 2x)^{-2} in ascending powers of xx, up to and including the term in x2x^{2}. (3 marks)
(b) Hence find the coefficient of x2x^{2} in the expansion of 2+3x(1−2x)2\dfrac{2 + 3x}{(1 - 2x)^{2}}. (2 marks)
●●●●●Level 45 marksStart
Expand (8+3x)−13(8 + 3x)^{-\frac{1}{3}} in ascending powers of xx, up to and including the term in x2x^{2}, and state the range of validity. (5 marks)
●●●●●Level 45 marksStart
(a) Find the first two terms, in ascending powers of xx, of the binomial expansion of

(1−x)−12(1 - x)^{-\frac{1}{2}}

(2 marks)
(b) Hence, for small values of xx, show that

tan⁡2x+3cos⁡4x≈A+Bx+Cx2\tan 2x + \frac{3}{\sqrt{\cos 4x}} \approx A + Bx + Cx^2

where AA, BB and CC are constants to be found.

(4 marks)
●●●●●Level 46 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Find the first three terms, in ascending powers of xx, of the binomial expansion of

(4+3x)−12(4 + 3x)^{-\frac{1}{2}}

writing each term in simplest form.

(4 marks)
(b) Using the answer to part (a) and using algebraic integration, estimate the value of

∫0122x4+3x dx\int_0^{\frac{1}{2}} \frac{2x}{\sqrt{4 + 3x}}\,\mathrm{d}x

giving your answer to 4 significant figures.

(4 marks)
●●●●●Level 48 marksStart
f(x)=2(1+2x)(1−x)f(x) = \dfrac{2}{(1 + 2x)(1 - x)}.
(a) Express f(x)f(x) in partial fractions. (3 marks)
(b) Hence show that, for small xx, f(x)≈2−2x+6x2f(x) \approx 2 - 2x + 6x^2. (6 marks)
(c) State the range of values of xx for which the full expansion is valid, justifying your answer. (3 marks)
●●●●●Level 512 marksStart
(a) Find, in ascending powers of xx, the first three non-zero terms of the binomial series expansion of

19+x∣x∣<9\dfrac{1}{\sqrt{9 + x}} \qquad |x| < 9

giving each coefficient as a simplified fraction.

(4 marks)

Using the expansion from part (a),
(b) state the first three non-zero terms of the binomial series expansion of 19−x\dfrac{1}{\sqrt{9 - x}}

(1 mark)
(c) Hence, or otherwise, show that

181−x2≈a+bx2\dfrac{1}{\sqrt{81 - x^2}} \approx a + bx^2

where aa and bb are fully simplified fractions to be found.

(2 marks)
(d) Use 181−x2≈a+bx2\dfrac{1}{\sqrt{81 - x^2}} \approx a + bx^2 with x=1x = 1 and the values of aa and bb to find a fully simplified rational approximation for 5\sqrt{5}

Show your working and make your method clear.

(3 marks)
●●●●●Level 510 marksStart
Find the first three terms, in ascending powers of xx, of (4+x)1/2(4 + x)^{1/2}, and state the range of validity.
●●●●●Level 54 marksStart
Given that 3(1+x)(1−2x)≡11+x+21−2x\dfrac{3}{(1 + x)(1 - 2x)} \equiv \dfrac{1}{1 + x} + \dfrac{2}{1 - 2x}, find the expansion of 3(1+x)(1−2x)\dfrac{3}{(1 + x)(1 - 2x)} up to and including the term in x2x^2, and state the range of validity.
●●●●●Level 54 marksStart
f(x)=7+x(1+x)(2−x)\mathrm{f}(x) = \frac{7 + x}{(1 + x)(2 - x)}
(a) Express f(x)\mathrm{f}(x) in partial fractions. (3 marks)
(b) Hence find the first three terms, in ascending powers of xx, of the expansion of f(x)\mathrm{f}(x). (4 marks)
(c) State the range of xx for which the full expansion is valid. (1 mark)
●●●●●Level 58 marksStart
(a) Expand (1−x)12(1 - x)^{\frac{1}{2}} in ascending powers of xx, up to and including the term in x3x^{3}. (3 marks)
(b) Substitute x=150x = \dfrac{1}{50} to evaluate the expansion, giving 88 decimal places. (2 marks)
(c) Using 1−150=752\sqrt{1 - \dfrac{1}{50}} = \dfrac{7}{5\sqrt{2}}, deduce an approximation to 2\sqrt{2}, giving your answer to 55 decimal places. (2 marks)
●●●●●Level 57 marksStart
In the expansion of (1+ax)n(1 + ax)^{n} in ascending powers of xx, the coefficient of xx is 1212 and the coefficient of x2x^{2} is 9090.
(a) Find the values of nn and aa, given that nn is a negative integer. (5 marks)
(b) Find the coefficient of x3x^{3}. (2 marks)
●●●●●Level 57 marksStart
f(x)=1+2x1−2x\mathrm{f}(x) = \sqrt{\frac{1 + 2x}{1 - 2x}}
(a) Find the series expansions of (1+2x)12(1 + 2x)^{\frac{1}{2}} and (1−2x)−12(1 - 2x)^{-\frac{1}{2}}, each up to and including the term in x2x^{2}. (4 marks)
(b) Hence show that f(x)≈1+2x+2x2\mathrm{f}(x) \approx 1 + 2x + 2x^{2} for small xx. (2 marks)
(c) Use x=0.1x = 0.1 to estimate 1.5\sqrt{1.5}, and state its value to 33 decimal places for comparison. (1 mark)
●●●●●Level 57 marksStart