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SEQUENCES AND SERIES C2 Worksheet A 1 For each of the following geometric series, write down the common ratio and find the value of
the eighth term.
a 3 + 9 + 27 + 81 + … b 1024 + 256 + 64 + 16 + … c 1 − 2 + 4 − 8 + … 2 For each of the following geometric series, find an expression for the nth term.
a 1 + 5 + 25 + 125 + … b 3 − 12 + 48 − 192 + … c 81 + 54 + 36 + 24 + … 3 Find the sum of the first 12 terms of each of the following geometric series.
a 2 + 4 + 8 + 16 + … b 640 + 320 + 160 + 80 + … c 16 − 1
2 + 121 − 1
24 + … 4 Given the first term, a, the common ratio, r, and the number of terms, n, find the sum of each of
the following geometric series. Give your answers to 3 decimal places where appropriate.
a a = 4, r = 3, n = 8 b a = 48, r = 12 , n = 14 c a = −1, r = −4, n = 12
d a = 200, r = 0.7, n = 20 e a = 120, r = 34− , n = 15 f a = −25, r = 1.2, n = 30
5 Evaluate to an appropriate degree of accuracy
a 9
13r
r =∑ b
61
18r
r
+
=∑ c
10
1(10 2 )r
r =×∑ d
8
1(0.8)r
r =∑
e 10
16
112 ( )r
r =
× ∑ f 9
1( 4)r
r =−∑ g
2012
4( )r
r=∑ h
9
32 ( 3)r
r=
× − ∑
6 The second and third terms of a geometric series are 2 and 10 respectively.
a Find the common ratio of the series.
b Find the first term of the series.
c Find the sum of the first eight terms of the series. 7 The first and fourth terms of a geometric series are 2 and 54 respectively.
a Find the common ratio of the series.
b Find the ninth term of the series. 8 The third and fourth terms of a geometric series are 24 and 8 respectively.
a Find the common ratio of the series.
b Find the first term of the series.
c Find, to 3 decimal places, the sum of the first 11 terms of the series. 9 The first and third terms of a geometric series are 6 and 24 respectively.
a Find the two possible values for the common ratio of the series.
Given also that the common ratio of the series is positive,
b find the sum of the first 15 terms of the series. 10 The first and fourth terms of a geometric series are 768 and −96 respectively.
a Find the common ratio of the series.
b Find the tenth term of the series.
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11 The second and fifth terms of a geometric series are 0.5 and 32 respectively.
a Find the first term and common ratio of the series.
b Find the number of terms of the series that are smaller than 10 000. 12 The sum of the first four terms of a geometric series is 130 and its common ratio is 1
21 .
a Find the first term of the series.
b Find the eighth term of the series.
c Find the least value of n for which the sum of the first n terms of the series is greater than 30 000.
13 All the terms of a geometric series are positive. The sum of the first and second terms of the
series is 10.8 and the sum of the third and fourth terms of the series is 43.2
a Find the first term and common ratio of the series.
b Find the sum of the first 16 terms of the series. 14 For each of the following geometric series, either find its sum to infinity or explain why this
cannot be found.
a 12 + 6 + 3 + 1.5 + … b 270 + 90 + 30 + 10 + … c 25 − 30 + 36 − 43.2 + …
d 216 + 144 + 96 + 64 + … e 825 + 2
5 + 12 + 5
8 + … f 500 − 300 + 180 − 108 + … 15 Find the sum to infinity of the geometric series with nth term
a (0.9)n b 6 × ( 12 )n c ( 3
4− )n − 1 d 40 × (0.8)n
16 A geometric series has first term 80 and common ratio 0.2
a Find the sum to infinity of the series.
b Find the difference between the sum to infinity of the series and the sum of the first six terms of the series.
17 A sequence is defined by the recurrence relation
un+1 = 13 un, n > 0, u1 = 1.
a Write down the first four terms of the sequence.
b Evaluate 1r
∞
=∑ ur .
18 The common ratio of a geometric series is 0.55 and the sum to infinity of the series is 40.
a Find the first term of the series.
b Find the smallest value of n for which the nth term of the series is less than 0.001 19 The sum, Sn , of the first n terms of a geometric series is given by Sn = 2n − 1.
a Find the first term and the fifth term of the series.
b Find an expression for the nth term of the series. 20 The first three terms of a geometric series are (k + 10), k and (k − 6) respectively.
a Find the value of the constant k.
b Find the sum to infinity of the series.
C2 SEQUENCES AND SERIES Worksheet A continued
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SEQUENCES AND SERIES C2 Worksheet B 1 The third and fourth terms of a geometric series are 27 and 1
420 respectively.
a Find the first term of the series.
b Find the sum to infinity of the series. 2 The first three terms of a geometric series are (k − 8), (k + 4) and (3k + 2) respectively, where k is
a positive constant.
a Find the value of k.
b Find the sixth term of the series.
c Show that the sum of the first ten terms of the series is 50 857.3 to 1 decimal place. 3 The second and fifth terms of a geometric series are 75 and 129.6 respectively.
a Show that the first term of the series is 62.5
b Find the value of the tenth term of the series to 1 decimal place.
c Find the sum of the first 12 terms of the series to 1 decimal place. 4 a Prove that the sum, Sn , of the first n terms of a geometric series with first term a and
common ratio r is given by
Sn = (1 )1
na rr
−−
.
b A geometric series has first term 2 and common ratio 2 .
Given that the sum of the first n terms of the series is 126( 2 + 1), find the value of n. 5 The first term of a geometric series is 18 and the sum to infinity of the series is 15.
a Find the common ratio of the series.
b Find the third term of the series.
c Find the exact difference between the sum of the first eight terms of the series and the sum to infinity of the series.
6 The sum of the first n terms of a geometric series is given by 5(3n − 1).
a Show that the third term of the series is 90.
b Find an expression for the nth term of the series in the form k(3n) where k is an exact fraction. 7 A student programs a computer to draw a series of straight lines with each line beginning at the
end of the previous one and at right angles to it. The first line is 4 mm long and thereafter each line is 25% longer than the previous one, so that a spiral is formed as shown above.
a Find the length, in mm, of the eighth straight line drawn by the program.
b Find the total length of the spiral, in metres, when 20 straight lines have been drawn.
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8 The second and fourth terms of a geometric series are 30 and 2.7 respectively.
Given that the common ratio, r, of the series is positive,
a find the value of r and the first term of the series,
b find the sum to infinity of the series.
9 a Evaluate 10
33r
r=∑ .
b Show that 15
1(2 12 )r
rr
=−∑ = 64 094.
10 A geometric series has common ratio r and the nth term of the series is denoted by un.
Given that u1 = 64 and that u3 − u2 = 20,
a show that 16r2 − 16r − 5 = 0,
b find the two possible values of r,
c find the fourth term of the series corresponding to each possible value of r.
d Taking the value of r such that the series converges, find the sum to infinity of the series. 11 A geometric series has first term 4 and common ratio 1
2 .
a Find the eighth term of the series as an exact fraction.
b Find the nth term of the series in the form 2y where y is a function of n.
c Show that the sum of the first n terms of the series is 8 − 23 − n. 12 The sequence of terms u1 , u2 , u3 , … is defined by
un = 4 × 3n, n ≥ 1.
a Find u6 .
b Find the smallest value of t such that the sum of the first t terms of the sequence is greater than 1025.
13 The sum of the first and third terms of a geometric series is 150. The sum of the second and
fourth terms of the series is −75.
a Find the first term and common ratio of the series.
b Find the sum to infinity of the series. 14 Three consecutive terms of an arithmetic series are a, b and (3a + 4) respectively.
a Find an expression for b in terms of a.
Given also that a, b and (6a + 1) respectively are consecutive terms of a geometric series and that a and b are integers,
b find the values of a and b. 15 When a ball is dropped onto a horizontal floor it bounces such that it reaches a maximum height
of 60% of the height from which it was dropped.
a Find the maximum height the ball reaches after its fourth bounce when it is initially dropped from 3 metres above the floor.
b Show that when the ball is dropped from a height of h metres above the floor it travels a total distance of 4h metres before coming to rest.
C2 SEQUENCES AND SERIES Worksheet B continued
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SEQUENCES AND SERIES C2 Worksheet C 1 Expand each of the following, simplifying the coefficient in each term.
a (1 + x)4 b (1 – x)5 c (1 + 4x)3 d (1 – 2y)3
e (1 + 12 x)4 f (1 + 1
3 y)3 g (1 + x2)5 h (1 – 32 x)4
2 Expand each of the following, simplifying the coefficient in each term.
a (x + y)3 b (a – b)5 c (x + 2y)4 d (2 + y)3
e (3 – x)3 f (5 + 2x)4 g (3 – 4y)5 h (3 + 12 x)4
3 Find the first four terms in the expansion in ascending powers of x of
a (1 + x)10 b (1 – x)6 c (1 + 2x)8 d (1 – 12 x)7
e (1 + x3)6 f (2 + x)9 g (3 – x)7 h (2 + 5x)10 4 Find the coefficient indicated in the following expansions.
a (1 + x)20, coefficient of x3 b (1 – x)14, coefficient of x4
c (1 + 4x)9, coefficient of x2 d (1 – 3y)14, coefficient of y3
e (1 – 13 x)12, coefficient of x4 f (1 − 1
2 x)16, coefficient of x5
g (1 + 25 x)15, coefficient of x2 h (1 + y2)8, coefficient of y6
5 Express each of the following in the required form where a and b are integers.
a (1 + 5 )3 in the form a + b 5 b (1 – 3 )4 in the form a + b 3
c (2 + 2 )3 in the form a + b 2 d (1 + 2 3 )4 in the form a + b 3 6 a Expand (1 + x)6 in ascending powers of x up to and including the term in x3, simplifying each coefficient.
b By substituting a suitable value of x into your answer for part a, obtain an estimate for
i 1.026 ii 0.996
giving your answers to 4 decimal places. 7 a Expand (1 + 2y)8 in ascending powers of y up to and including the term in y3, simplifying each coefficient.
b By substituting a suitable value of y into your answer for part a, obtain an estimate for
i 0.988 ii 1.018
giving your answers to 4 decimal places. 8 Expand and simplify
a (1 + x)4 + (1 – x)4 b (1 – 13 x)3 – (1 + 1
3 x)3 9 The coefficient of x2 in the expansion of (1 + ax)4 in ascending powers of x is 24, where a is
a constant and a < 0. Find
a the value of a,
b the value of the coefficient of x3 in the expansion.
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SEQUENCES AND SERIES C2 Worksheet D 1 Expand
a (1 + 3x)4 b (2 – x)5 c (3 + 10x2)3 d (a + 2b)5
e (x2 – y)3 f (5 + 12 x)4 g (x + 1
x)4 h (t – 2
2t
)3
2 Find the first four terms in the expansion in ascending powers of x of
a (1 + 3x)6 b (1 – 14 x)8 c (5 – x)7 d (3 + 2x2)10
3 Find the coefficient indicated in the following expansions
a (1 + x)15, coefficient of x3 b (1 – 2x)12, coefficient of x4
c (3 + x)7, coefficient of x2 d (2 – y)10, coefficient of y5
e (2 + t3)8, coefficient of t15 f (1 − 1x
)9, coefficient of x–3
4 a Express ( 2 – 5 )4 in the form a + b 10 , where a, b ∈ .
b Express ( 2 + 13
)3 in the form a 2 + b 3 , where a, b ∈ .
c Express (1 + 5 )3 – (1 – 5 )3 in the form a 5 , where a ∈ .
5 a Expand (1 + 2x )10 in ascending powers of x up to and including the term in x3, simplifying
each coefficient.
b By substituting a suitable value of x into your answer for part a, obtain an estimate for
i 1.00510 ii 0.99610
giving your answers to 5 decimal places. 6 a Expand (3 + x)8 in ascending powers of x up to and including the term in x3, simplifying
each coefficient.
b By substituting a suitable value of x into your answer for part a, obtain an estimate for
i 3.0018 ii 2.9958
giving your answers to 7 significant figures. 7 Expand and simplify
a (1 + 10x)4 + (1 – 10x)4 b (2 – 13 x)3 – (2 + 1
3 x)3
c (1 + 4y)(1 + y)3 d (1 – x)(1 + 1x
)3
8 Expand each of the following in ascending powers of x up to and including the term in x3.
a (1 + x2)(1 – 3x)10 b (1 – 2x)(1 + x)8
c (1 + x + x2)(1 – x)6 d (1 + 3x – x2)(1 + 2x)7 9 Find the term independent of y in each of the following expansions.
a (y + 1y
)8 b (2y – 12y
)12 c ( 1y
+ y2)6 d (3y − 21y
)9
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10 The coefficient of x2 in the binomial expansion of (1 + 2
5 x)n, where n is a positive integer, is 1.6
a Find the value of n.
b Use your value of n to find the coefficient of x4 in the expansion. 11 Given that y1 = (1 – 2x)(1 + x)10 and y2 = ax2 + bx + c and that when x is small, y2 can be used
as an approximation for y1,
a find the values of the constants a, b and c,
b find the percentage error in using y2 as an approximation for y1 when x = 0.2 12 In the binomial expansion of (1 + px)q, where p and q are constants and q is a positive integer, the
coefficient of x is –12 and the coefficient of x2 is 60.
Find
a the value of p and the value of q,
b the value of the coefficient of x3 in the expansion. 13 a Expand (3 –
3x )12 as a binomial series in ascending powers of x up to and including the term
in x3, giving each coefficient as an integer.
b Use your series expansion with a suitable value of x to obtain an estimate for 2.99812, giving your answer to 2 decimal places.
14 a Expand (1 – x)5 as a binomial series in ascending powers of x.
b Express ( 3 + 1)( 3 – 2) in the form A + B 3 , where A, B ∈ .
c Hence express each of the following in the form C + D 3 , where C, D ∈ .
i ( 3 + 1)5( 3 – 2)5
ii ( 3 + 1)6( 3 – 2)5
15 a Expand (1 + 2x )9 in ascending powers of x up to and including the term in x4.
Hence, or otherwise, find
b the coefficient of x3 in the expansion of (1 + 2x )9 – (1 –
2x )9,
c the coefficient of x4 in the expansion of (1 + 2x)(1 + 2x )9.
16 The term independent of x in the expansion of (x3 + 2ax
)5 is –80.
Find the value of the constant a.
17 In the binomial expansion of (1 + xk
)n, where k is a non-zero constant, n is an integer and n > 1,
the coefficient of x2 is three times the coefficient of x3.
a Show that k = n – 2.
Given also that n = 7,
b expand (1 + xk
)n in ascending powers of x up to and including the term in x4, giving each
coefficient as a fraction in its simplest form.
C2 SEQUENCES AND SERIES Worksheet D continued
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SEQUENCES AND SERIES C2 Worksheet E 1 Expand (1 + 4x)4 in ascending powers of x, simplifying the coefficients. (4) 2 A geometric series has first term 3 and common ratio −2.
a Find the fifth term of the series. (2)
b Find the sum of the first ten terms of the series. (2)
c Show that the sum of the first eight positive terms of the series is 65 535. (4) 3 a Expand (1 + 3x)7 in ascending powers of x up to and including the term in x4,
simplifying each coefficient in the expansion. (4)
b Use your series with a suitable value of x to estimate the value of 1.037 correct to 5 decimal places. (3)
4 Evaluate 12
32r
r=∑ . (4)
5 a Expand (2 + x)5, simplifying the coefficient in each term. (4)
b Hence, or otherwise, write down the expansion of (2 – x)5. (1)
c Show that
(2 + 5 )5 – (2 – 5 )5 = k 5 ,
where k is an integer to be found. (4) 6 Ginny opens a savings account and decides to pay £200 into the account at the start of
each month. At the end of each month, interest of 0.5% is paid into the account.
a Find, to the nearest penny, the interest paid into the account at the end of the third month. (4)
b Show that the total interest paid into the account over the first 12 months is £79.45 to the nearest penny. (5)
7 Find the first four terms in the expansion of (1 − 3x)8 in ascending powers of x,
simplifying each coefficient. (4) 8 a Prove that the sum, Sn , of the first n terms of a geometric series with first term a and common ratio r is given by
Sn = (1 )1
na rr
−−
. (4)
b Find the exact sum of the first 16 terms of the geometric series with fourth term 3 and fifth term 6. (5)
9 a Write down the first three terms in the binomial expansion of (1 + ax)n, where n is a
positive integer, in ascending powers of x. (2)
Given that the coefficient of x2 is three times the coefficient of x,
b show that n = a
a+6 . (4)
Given also that a = 23 ,
c find the coefficient of x3 in the expansion. (3)
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10 Find the first three terms in the expansion of (2 + 5x)6 in ascending powers of x,
simplifying each coefficient. (4) 11 The first term of a geometric series is 162 and the sum to infinity of the series is 486.
a Find the common ratio of the series. (3)
b Find the sixth term of the series. (2)
c Find, to 3 decimal places, the sum of the first ten terms of the series. (4) 12 a Expand (1 + 3x)4 in ascending powers of x, simplifying the coefficients. (4)
b Find the coefficient of x2 in the expansion of
(1 + 4x − x2)(1 + 3x)4. (3) 13 In a computer game, each player must complete the tasks set at each level within a
fixed amount of time in order to progress to the next level.
The time allowed for level 1 is 2 minutes and the time allowed for each of the other levels is 10% less than that allowed in the previous level.
a Find, in seconds, the time allowed for completing level 4. (2)
b Find, in minutes and seconds, the maximum total time allowed for completing the first 12 levels of the game. (4)
14 Given that (1 +
2x )8(1 – x)6 ≡ 1 + Ax + Bx2 + … ,
find the values of the constants A and B. (7) 15 The terms of a sequence are defined by the recurrence relation
ur = 2ur–1, r > 1, u1 = 6.
a Write down the first four terms of the sequence. (1)
b Evaluate 10
1r=∑ ur . (3)
16 a Expand (1 + x)4 in ascending powers of x. (2)
b Hence, or otherwise, write down the expansion of (1 – x)4 in ascending powers of x. (1)
c By using your answers to parts a and b, or otherwise, solve the equation
(1 + x)4 + (1 – x)4 = 82,
for real values of x. (5) 17 The common ratio of a geometric series is 1.5 and the third term of the series is 18.
a Find the first term of the series. (2)
b Find the sum of the first six terms of the series. (2)
c Find the smallest value of k such that the kth term of the series is greater than 8000. (4)
18 The first two terms in the expansion of (1 + 2ax )10 + (1 + bx)10, in ascending powers of x,
are 2 and 90x2.
Given that a < b, find the values of the constants a and b. (9)
C2 SEQUENCES AND SERIES Worksheet E continued
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SEQUENCES AND SERIES C2 Worksheet F 1 The first and fourth terms of a geometric series are 108 and 32 respectively.
a Find the third term of the series. (5)
b Find the sum to infinity of the series. (2) 2 Expand (1 − 2x)5 in ascending powers of x, simplifying each coefficient. (4) 3 An internet site has 3600 subscribers at the start of a promotional campaign.
In a model of the results of the campaign, it is assumed that the site will gain 200 new subscribers in the first week and that in subsequent weeks the number of new subscribers will be 15% greater each week.
a Show that, according to this model, the site will gain 304 new subscribers in the fourth week of the campaign. (2)
b Find the total number of subcribers to the site predicted by the model after ten weeks of the campaign, assuming that no subscriptions are cancelled in this period. (5)
4 a Find the first three terms in the expansion of (1 + 4x)7 in ascending powers of x. (3)
b Hence, find the coefficient of x2 in the expansion of
(1 + 2x)2(1 + 4x)7. (3)
5 a Write down the first four terms in the expansion in ascending powers of x of (1 + xk
)2n,
where k is a non-zero constant, n is an integer and n > 1. (4)
Given that the coefficient of x3 is half the coefficient of x2,
b show that 3k = 4(n – 1). (3)
Given also that the coefficient of x is 2,
c find the values of n and k. (3) 6 The second and third terms of a geometric series are 6 and 3 2 respectively.
a Find, in surd form, the first term and the common ratio of the series. (4)
b Show that the sum of the first eight terms of the series is 40 2 ( 3 + 1). (4)
7 Evaluate 9
1(3 1)r
r=−∑ . (5)
8 a Find the first four terms in the expansion of (1 + 2x)9 in ascending powers of x. (4)
b Show that, if terms involving x4 and higher powers of x may be ignored,
(1 + 2x)9 + (1 − 2x)9 = 2 + 288x2. (3)
c Hence find the value of
1.0029 + 0.9989,
giving your answer to 7 significant figures. (2) 9 Given that
(k – x)9 ≡ a – bx + bx2 + …,
find the values of the positive integers a, b and k. (7)
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10 Expand (3 + 2x)4 in ascending powers of x, simplifying the coefficients. (4) 11 The first term of a geometric series is t and the sum to infinity of the series is 3t.
a Find the common ratio of the series. (3)
Given also that the sum of the first four terms of the series is 130,
b find the value of t. (4) 12 a Expand (1 − 2x)4 in ascending powers of x, simplifying the coefficients. (4)
b Hence, or otherwise, find the coefficient of y2 in the expansion of
(1 + 4y − 2y2)4. (4) 13 A company buys a new car for £12 000 at the start of one year. In a model, it is assumed
that each year the value of a car decreases by 25% of its value at the start of that year.
a Show that the value of the car after four years is £3800 to 3 significant figures. (2)
The company plans to buy one new car for £12 000 at the start of each subsequent year.
b Using the same model, find the total value of all the cars the company will have bought under this plan immediately after the purchase of the eighth car. (5)
14 The polynomial p(x) is defined by
p(x) = (x + 3)4 − (x + 1)4.
a Show that (x + 2) is a factor of p(x). (2)
b Fully factorise p(x). (7)
c Hence show that there is only one real solution to the equation p(x) = 0. (3) 15 f(x) ≡ (1 – x)(1 + 2x)n, n ∈ .
Given that the coefficient of x2 in the binomial expansion of f(x) is 198, find
a the value of n, (6)
b the coefficient of x3 in the expansion. (3)
16 Expand ( 3x
– x)4 in descending powers of x, simplifying the coefficient in each term. (4)
17 The sum, Sn , of the first n terms of a series is given by
Sn = 3n − 1.
a Show that the fourth term of the series is 54. (3)
b Show that the nth term of the series can be expressed in the form k(3n) where k is an exact fraction to be found. (4)
c Prove that the series is geometric. (3) 18 An arithmetic series has first term 3, second term x and fourth term y.
a Find an expression for y in terms of x. (3)
Given also that 3, x and y are the first, second and fourth terms respectively of a geometric series,
b show that x3 − 27x + 54 = 0, (4)
c by first finding a linear factor of x3 − 27x + 54, find the two possible values of x. (6)
C2 SEQUENCES AND SERIES Worksheet F continued
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SEQUENCES AND SERIES C2 Answers - Worksheet A 1 a r = 3 b r = 1
4 c r = −2
u8 = 3 × 37 = 6561 u8 = 1024 × 71
4( ) = 116 u8 = 1 × (−2)7 = −128
2 a a = 1, r = 5 b a = 3, r = −4 c a = 81, r = 2
3
un = 5n − 1 un = 3 × (−4)n − 1 un = 81 × 12
3( )n − 3 a a = 2, r = 2, n = 12 b a = 640, r = 1
2 , n = 12 c a = 16 , r = −3, n = 12
S12 = 122(2 1)
2 1−
− = 8190 S12 =
1212
12
640[1 ( ) ]1
−
−= 11
161279 S12 = 121
6[1 ( 3) ]1 ( 3)
− −− −
= 1322143−
4 a S8 = 84(3 1)
3 1−
− = 13 120 b S14 =
141212
48[1 ( ) ]
1−
− = 95.994 c S12 =
12[1 ( 4) ]1 ( 4)
− − −− −
= 3 355 443
d S20 = 20200[1 (0.7) ]
1 0.7−−
e S15 = 153
434
120[1 ( ) ]
1 ( )
− −
− − f S30 =
3025[(1.2) 1]1.2 1
− −−
= 666.135 = 69.488 = −29 547.039 5 a GP: a = 3 b GP: a = 64 c GP: a = 20 d GP: a = 0.8 r = 3, n = 9 r = 8, n = 6 r = 2, n = 10 r = 0.8, n = 8
S9 = 93(3 1)
3 1−
− S6 =
664(8 1)8 1
−−
S10 = 1020(2 1)
2 1−
− S8 =
80.8[1 (0.8) ]1 0.8
−−
= 29 523 = 2 396 736 = 20 460 = 3.329 (3dp)
e GP: a = 2 f GP: a = −4 g GP: a = 116 h GP: a = −54
r = 16 , n = 10 r = −4, n = 9 r = 1
2 , n = 17 r = −3, n = 7
S10 = 101
616
2[1 ( ) ]
1
−
− S9 =
94[1 ( 4) ]1 ( 4)
− − −− −
S17 = 171 1
16 212
[1 ( ) ]
1
−
− S7 =
754[1 ( 3) ]1 ( 3)
− − −− −
= 2.400 (3dp) = −209 716 = 0.125 (3dp) = −29 538 6 a r = 10 ÷ 2 = 5 7 a a = 2, ar3 = 54 ∴ r3 = 54 ÷ 2 = 27 b a × 5 = 2 ∴ a = 0.4 r = 3 27 = 3
c S8 = 80.4(5 1)
5 1−
− = 39 062.4 b u9 = 2 × 38 = 13 122
8 a r = 8 ÷ 24 = 13 9 a a = 6, ar2 = 24 ∴ r2 = 24 ÷ 6 = 4
b a × ( 13 )2 = 24 ∴ a = 216 r = ± 2
c S11 = 111
313
216[1 ( ) ]
1
−
− = 323.998 b r = 2, S15 =
156(2 1)2 1
−−
= 196 602
10 a a = 768, ar3 = −96 11 a ar = 0.5, ar4 = 32 ∴ r3 = 32 ÷ 0.5 = 64
r3 = −96 ÷ 768 = 18− r = 3 64 = 4, a × 4 = 0.5 ∴ a = 0.125
r = 138− = 1
2− b 0.125 × 4n − 1 < 10 000 ∴ 4n − 1 < 80 000
b u10 = 768 × ( 12− )9 = −1.5 (n − 1) lg 4 < lg 80 000
n < lg80000lg 4
+ 1
n < 9.14 ∴ 9 terms
C2 SEQUENCES AND SERIES Answers - Worksheet A page 2
Solomon Press
12 a 43
232
[( ) 1]
1
a −
− = 130 13 a a + ar = a(1 + r) = 10.8
a = 130 ÷ 658 = 16 ar2 + ar3 = ar2(1 + r) = 43.2
b u8 = 16 × ( 32 )7 = 3
8273 ∴ r2 = 43.2 ÷ 10.8 = 4
c 3232
16[( ) 1]
1
n −
− > 30 000 all terms +ve ∴ r +ve ∴ r = 2
( 32 )n > 938.5 sub. a = 10.8 ÷ 3 = 3.6
n lg 32 > lg 938.5 b S16 =
163.6(2 1)2 1
−−
= 235 926
n > lg938.5lg1.5
n > 16.9 ∴ least n = 17 14 a a = 12, r = 0.5 b a = 270, r = 1
3 c a = 25, r = −1.2
S∞ = 121 0.5−
= 24 S∞ = 13
2701−
= 405 no S∞ as r < −1 ∴ diverges
d a = 216, r = 23 e a = 8
25 , r = 54 f a = 500, r = −0.6
S∞ = 23
2161−
= 648 no S∞ as r > 1 ∴ diverges S∞ = 5001 ( 0.6)− −
= 312.5
15 a a = 0.9, r = 0.9 b a = 3, r = 1
2 c a = 1, r = 34− d a = 32, r = 0.8
S∞ = 0.91 0.9−
= 9 S∞ = 12
31−
= 6 S∞ = 34
11 ( )− −
= 47 S∞ = 32
1 0.8− = 160
16 a S∞ = 80
1 0.2− = 100 17 a 1, 1
3 , 19 , 1
27
b S6 = 680[1 (0.2) ]
1 0.2−−
= 99.9936 b GP: a = 1, r = 13
S∞ − S6 = 0.0064 S∞ = 13
11−
= 32
18 a
1 0.55a
− = 40 19 a u1 = S1 = 21 − 1 = 1
a = 0.45 × 40 = 18 S5 = 25 − 1 = 31, S4 = 24 − 1 = 15 b 18 × (0.55)n − 1 < 0.001 u5 = S5 − S4 = 31 − 15 = 16 (n − 1) lg 0.55 < lg 0.0000556 b Sn − 1 = 2n − 1 − 1 n > lg0.0000556
lg0.55 + 1 un = Sn − Sn − 1 = (2n − 1) − (2n − 1 − 1)
n > 17.4 ∴ smallest n = 18 = 2n − 2n − 1 = 2n − 1(2 − 1) = 2n − 1 20 a
10k
k + = 6k
k−
k2 = (k + 10)(k − 6) 4k − 60 = 0 k = 15 b u1 = 25, u2 = 15 ∴ a = 25, r = 0.6 S∞ = 25
1 0.6− = 62.5
Solomon Press
SEQUENCES AND SERIES C2 Answers - Worksheet B
1 a r = 1
420 ÷ 27 = 34 2 a 4
8kk
+−
= 3 24
kk
++
a × ( 34 )2 = 27 (k + 4)2 = (3k + 2)(k − 8)
a = 169 × 27 = 48 k2 − 15k − 16 = 0
b S∞ = 34
481−
= 192 (k + 1)(k − 16) = 0
k > 0 ∴ k = 16 b u1 = 8, u2 = 20 ∴ a = 8, r = 5
2 u6 = 8 × ( 5
2 )5 = 14781
c S10 = 105
252
8[( ) 1]
1
−
− = 50 857.3
3 a ar = 75, ar4 = 129.6 4 a Sn = a + ar + ar2 + … arn − 2 + arn − 1 r3 = 129.6 ÷ 75 = 1.728 rSn = ar + ar2 + ar3 + … arn − 1 + arn r = 3 1.728 = 1.2 subtracting, a = 75 ÷ 1.2 = 62.5 Sn − rSn = a − arn b u10 = 62.5 × (1.2)9 = 322.5 (1 − r)Sn = a(1 − rn)
c S12 = 1262.5[(1.2) 1]
1.2 1−
− = 2473.8 Sn = (1 )
1
na rr
−−
b 2[1 ( 2) ]1 2
n−−
= 126( 2 + 1)
1 − ( 2 )n = 63( 2 + 1)(1 − 2 ) 1 − ( 2 )n = 63(1 − 2) ( 2 )n = 64
122 n = 26
n = 12 5 a 18
1 r− = 15 6 a S3 = 5(33 − 1) = 130
∴ 1 − r = 1815 = 1.2 S2 = 5(32 − 1) = 40
r = −0.2 u3 = S3 − S2 = 90 b u3 = 18 × (−0.2)2 = 0.72 b Sn − 1 = 5(3n − 1 − 1)
c S8 = 818[1 ( 0.2) ]
1 ( 0.2)− −− −
= 14.9999616 un = Sn − Sn − 1 = 5(3n − 1) − 5(3n − 1 − 1)
S∞ − S8 = 0.000 0384 = 5[3n − 3n − 1] = 5(3n)[1 − 13 ] = 10
3 (3n)
7 a 4 × (1.25)7 = 19.1 mm (3sf) 8 a ar = 30, ar3 = 2.7 ∴ r2 = 2.7 ÷ 30 = 0.09 b GP: a = 4, r = 1.25 r > 0 ∴ r = 0.09 = 0.3
S20 = 204[(1.25) 1]
1.25 1−
− = 1371.8 mm a = 30 ÷ 0.3 = 100
∴ length = 1.37 m (3sf) b S∞ = 1001 0.3−
= 142.9 (1dp)
C2 SEQUENCES AND SERIES Answers - Worksheet B page 2
Solomon Press
9 a GP: a = 27, r = 3 10 a a = 64, ar2 − ar = 20
S8 = 827(3 1)
3 1−
− = 88 560 ∴ 64r2 − 64r = 20
b 15
1r=∑ 2r : GP, a = 2, r = 2 16r2 − 16r − 5 = 0
S15 = 152(2 1)
2 1−
− = 65 534 b (4r + 1)(4r − 5) = 0
15
1r=∑ 12r : AP, a = 12, d = 12 r = 1
4− or 54
S15 = 152 [24 + (14 × 12)] = 1440 c r = 1
4− ⇒ u4 = 64 × ( 14− )3 = −1
15
1r=∑ (2r − 12r) = 65 534 − 1440 = 64 094 r = 5
4 ⇒ u4 = 64 × ( 54 )3 = 125
d r = 14− ⇒ S∞ = 1
4
641 ( )− −
= 1551
11 a u8 = 4 × ( 1
2 )7 = 132 12 a u6 = 4 × 36 = 2916
b un = 4 × ( 12 )n − 1 b GP: a = 12, r = 3
= 22 × 21 − n St = 12(3 1)3 1
t −−
= 6(3t − 1)
= 23 − n ∴ 6(3t − 1) > 1025
c Sn = 1212
4[1 ( ) ]1
n−
− 3t >
25106
+ 1
= 8(1 − 2−n) t lg 3 > lg (2510
6 + 1)
= 8 − (23 × 2−n) t > 25106lg( 1)
lg3+
= 8 − 23 − n t > 50.8 ∴ smallest t = 51 13 a a + ar2 = a(1 + r2) = 150 14 a b − a = (3a + 4) − b ar + ar3 = ar(1 + r2) = −75 2b = 4a + 4 ∴ r = −75 ÷ 150 = 1
2− b = 2a + 2
a = 150 ÷ 54 = 120 b 2 2a
a+ = 6 1
2 2aa
++
b S∞ = 12
1201 ( )− −
= 80 (2a + 2)2 = a(6a + 1)
2a2 − 7a − 4 = 0 (2a + 1)(a − 4) = 0 a integer ∴ a = 4 sub. b = 10 15 a after 4th bounce, reaches 3 × (0.6)4 = 0.3888 m b total distance = h + 2[0.6h + (0.6)2h + (0.6)3h + …] = h + 2 × S∞ of GP, a = 0.6h, r = 0.6 = h + 2 0.6
1 0.6h×
−
= h + 3h = 4h metres
Solomon Press
SEQUENCES AND SERIES C2 Answers - Worksheet C
1 a = 1 + 4x + 6x2 + 4x3 + x4 b = 1 − 5x + 10x2 − 10x3 + 5x4 − x5 c = 1 + 3(4x) + 3(4x)2 + (4x)3 d = 1 + 3(−2y) + 3(−2y)2 + (−2y)3 = 1 + 12x + 48x2 + 64x3 = 1 − 6y + 12y2 − 8y3
e = 1 + 4( 12 x) + 6( 1
2 x)2 + 4( 12 x)3 + ( 1
2 x)4 f = 1 + 3( 13 y) + 3( 1
3 y)2 + ( 13 y)3
= 1 + 2x + 32 x2 + 1
2 x3 + 116 x4 = 1 + y + 1
3 y2 + 127 y3
g = 1 + 5(x2) + 10(x2)2 + 10(x2)3 + 5(x2)4 + (x2)5 h = 1 + 4(− 32 x) + 6(− 3
2 x)2 + 4(− 32 x)3 + (− 3
2 x)4 = 1 + 5x2 + 10x4 + 10x6 + 5x8 + x10 = 1 − 6x + 27
2 x2 − 272 x3 + 81
16 x4
2 a = x3 + 3x2y + 3xy2 + y3 b = a5 − 5a4b + 10a3b2 − 10a2b3 + 5ab4 − b5 c = x4 + 4x3(2y) + 6x2(2y)2 + 4x(2y)3 + (2y)4 d = 23 + 3(22)y + 3(2)y2 + y3 = x4 + 8x3y + 24x2y2 + 32xy3 + 16y4 = 8 + 12y + 6y2 + y3 e = 33 + 3(32)(−x) + 3(3)(−x)2 + (−x)3 f = 54 + 4(53)(2x) + 6(52)(2x)2 + 4(5)(2x)3 + (2x)4 = 27 − 27x + 9x2 − x3 = 625 + 1000x + 600x2 + 160x3 + 16x4 g = 35 + 5(34)(−4y) + 10(33)(−4y)2 + 10(32)(−4y)3 + 5(3)(−4y)4 + (−4y)5 = 243 − 1620y + 4320y2 − 5760y3 + 3840y4 − 1024y5
h = 34 + 4(33)( 12 x) + 6(32)( 1
2 x)2 + 4(3)( 12 x)3 + ( 1
2 x)4 = 81 + 54x + 27
2 x2 + 32 x3 + 1
16 x4
3 a = 1 + 10x + 10 9
2× x2 + 10 9 8
3 2× ×× x3 + … b = 1 + 6(−x) + 6 5
2× (−x)2 + 6 5 4
3 2× ×
× (−x)3 + … = 1 + 10x + 45x2 + 120x3 + … = 1 − 6x + 15x2 − 20x3 + …
c = 1 + 8(2x) + 8 72× (2x)2 + 8 7 6
3 2× ×
× (2x)3 + … d = 1 + 7(− 12 x) + 7 6
2× (− 1
2 x)2 + 7 6 53 2× ×× (− 1
2 x)3 + … = 1 + 16x + 112x2 + 448x3 + … = 1 − 7
2 x + 214 x2 − 35
8 x3 + …
e = 1 + 6(x3) + 6 52× (x3)2 + 6 5 4
3 2× ×× (x3)3 + … f = 29 + 9(28)x + 9 8
2× (27)x2 + 9 8 7
3 2× ×× (26)x3 + …
= 1 + 6x3 + 15x6 + 20x9 + … = 512 + 2304x + 4608x2 + 5376x3 + …
g = 37 + 7(36)(−x) + 7 62× (35)(−x)2 + 7 6 5
3 2× ×× (34)(−x)3 + …
= 2187 − 5103x + 5103x2 − 2835x3 + …
h = 210 + 10(29)(5x) + 10 92× (28)(5x)2 + 10 9 8
3 2× ×× (27)(5x)3 + …
= 1024 + 25 600x + 288 000x2 + 1 920 000x3 + … 4 a = 20
3
= 1140 b = 144
× (−1)4 = 1001
c = 92
× 42 = 576 d = 143
× (−3)3 = −9828
e = 124
× (− 13 )4 = 55
9 or 196 f = 16
5
× (− 12 )5 = −136.5
g = 152
× ( 25 )2 = 84
5 or 16.8 h = 83
= 56
C2 SEQUENCES AND SERIES Answers - Worksheet C page 2
Solomon Press
5 a = 1 + 3( 5 ) + 3( 5 )2 + ( 5 )3 b = 1 + 4(− 3 ) + 6(− 3 )2 + 4(− 3 )3 + (− 3 )4 = 1 + 3 5 + 15 + 5 5 = 1 − 4 3 + 18 − 12 3 + 9 = 16 + 8 5 = 28 − 16 3
c = 23 + 3(22)( 2 ) + 3(2)( 2 )2 + ( 2 )3 d = 1 + 4(2 3 ) + 6(2 3 )2 + 4(2 3 )3 + (2 3 )4 = 8 + 12 2 + 12 + 2 2 = 1 + 8 3 + 72 + 96 3 + 144 = 20 + 14 2 = 217 + 104 3
6 a = 1 + 6x + 6 5
2× x2 + 6 5 4
3 2× ×× x3 + …
= 1 + 6x + 15x2 + 20x3 + …
b i let x = 0.02 1.026 ≈ 1 + 6(0.02) + 15(0.02)2 + 20(0.02)3 = 1 + 0.12 + 0.0060 + 0.000 160 = 1.1262 (4dp) ii let x = −0.01 0.996 ≈ 1 + 6(−0.01) + 15(−0.01)2 + 20(−0.01)3 = 1 − 0.06 + 0.0015 − 0.000 020 = 0.9415 (4dp)
7 a = 1 + 8(2y) + 8 7
2× (2y)2 + 8 7 6
3 2× ×
× (2y)3 + … = 1 + 16y + 112y2 + 448y3 + …
b i let y = −0.01 0.988 ≈ 1 + 16(−0.01) + 112(−0.01)2 + 448(−0.01)3 = 1 − 0.16 + 0.0112 − 0.000 448 = 0.8508 (4dp) ii let y = 0.005 1.018 ≈ 1 + 16(0.005) + 112(0.005)2 + 448(0.005)3 = 1 + 0.080 + 0.002 800 + 0.000 056 000 = 1.0829 (4dp)
8 a = 1 + 4x + 6x2 + 4x3 + x4 + (1 − 4x + 6x2 − 4x3 + x4) = 2 + 12x2 + 2x4
b = 1 + 3(− 13 x) + 3(− 1
3 x)2 + (− 13 x)3 − [1 + 3( 1
3 x) + 3( 13 x)2 + ( 1
3 x)3] = 1 − x + 1
3 x2 − 127 x3 − (1 + x + 1
3 x2 + 127 x3)
= −2x − 227 x3
9 a 6(ax)2 = 24x2 a2 = 4 a < 0 ∴ a = −2
b = 4a3 = −32
Solomon Press
SEQUENCES AND SERIES C2 Answers - Worksheet D
1 a = 1 + 4(3x) + 6(3x)2 + 4(3x)3 + (3x)4 = 1 + 12x + 54x2 + 108x3 + 81x4
b = 25 + 5(24)(−x) + 10(23)(−x)2 + 10(22)(−x)3 + 5(2)(−x)4 + (−x)5 = 32 − 80x + 80x2 − 40x3 + 10x4 − x5
c = 33 + 3(32)(10x2) + 3(3)(10x2)2 + (10x2)3 = 27 + 270x2 + 900x4 + 1000x6
d = a5 + 5a4(2b) + 10a3(2b)2 + 10a2(2b)3 + 5a(2b)4 + (2b)5 = a5 + 10a4b + 40a3b2 + 80a2b3 + 80ab4 + 32b5
e = (x2)3 + 3(x2)2(−y) + 3(x2)(−y)2 + (−y)3 = x6 − 3x4y + 3x2y2 − y3
f = 54 + 4(53)( 12 x) + 6(52)( 1
2 x)2 + 4(5)( 12 x)3 + ( 1
2 x)4 = 625 + 250x + 75
2 x2 + 52 x3 + 1
16 x4
g = x4 + 4x3( 1x
) + 6x2( 1x
)2 + 4x( 1x
)3 + ( 1x
)4
= x4 + 4x2 + 6 + 24x
+ 41x
h = t3 + 3t2(− 22t
) + 3t(− 22t
)2 + (− 22t
)3
= t3 − 6 + 312t
− 68t
2 a = 1 + 6(3x) + 6 52× (3x)2 + 6 5 4
3 2× ×× (3x)3 + …
= 1 + 18x + 135x2 + 540x3 + …
b = 1 + 8(− 14 x) + 8 7
2× (− 1
4 x)2 + 8 7 63 2× ×
× (− 14 x)3 + …
= 1 − 2x + 74 x2 − 7
8 x3 + …
c = 57 + 7(56)(−x) + 7 62× (55)(−x)2 + 7 6 5
3 2× ×× (54)(−x)3 + …
= 78 125 − 109 375x + 65 625x2 − 21 875x3 + …
d = 310 + 10(39)(2x2) + 10 92× (38)(2x2)2 + 10 9 8
3 2× ×× (37)(2x2)3 + …
= 59 049 + 393 660x2 + 1 180 980x4 + 2 099 520x6 + …
3 a = 153
= 455 b = 124
× (−2)4 = 7920
c = 72
× 35 = 5103 d = 105
× 25 × (−1)5 = −8064
e = 85
× 23 = 448 f = 93
× (−1)3 = −84
4 a = ( 2 )4 + 4( 2 )3(− 5 ) + 6( 2 )2(− 5 )2 + 4( 2 )(− 5 )3 + (− 5 )4 = 4 − 8 10 + 60 − 20 10 + 25 = 89 − 28 10
b = ( 2 )3 + 3( 2 )2( 13
) + 3( 2 )( 13
)2 + ( 13
)3
= 2 2 + 2 3 + 2 + 19 3
= 3 2 + 199 3
C2 SEQUENCES AND SERIES Answers - Worksheet D page 2
Solomon Press
c = 1 + 3( 5 ) + 3( 5 )2 + ( 5 )3 − [1 + 3(− 5 ) + 3(− 5 )2 + (− 5 )3] = 1 + 3 5 + 15 + 5 5 − [1 − 3 5 + 15 − 5 5 ] = 16 + 8 5 − [16 − 8 5 ] = 16 5
5 a = 1 + 10(
2x ) + 10 9
2× (
2x )2 + 10 9 8
3 2× ×× (
2x )3 + …
= 1 + 5x + 454 x2 + 15x3 + …
b i let x = 0.01 1.00510 ≈ 1 + 0.05 + 0.001 125 + 0.000 015 = 1.051 14 (5dp) ii let x = −0.008 0.99610 ≈ 1 − 0.040 + 0.000 720 − 0.000 007 680 = 0.960 71 (5dp) 6 a = 38 + 8(37)x + 8 7
2× (36)x2 + 8 7 6
3 2× ×
× (35)x3 + … = 6561 + 17 496x + 20 412x2 + 13 608x3 + …
b i let x = 0.001 3.0018 ≈ 6561 + 17.496 + 0.020 412 + 0.000 013 608 = 6578.516 (7sf) ii let x = −0.005 2.9958 ≈ 6561 − 87.480 + 0.510 300 − 0.001 701 000 = 6474.029 (7sf) 7 a (1 + 10x)4 = 1 + 4(10x) + 6(10x)2 + 4(10x)3 + (10x)4 = 1 + 40x + 600x2 + 4000x3 + 10 000x4 ∴ (1 + 10x)4 + (1 − 10x)4 = 1 + 40x + 600x2 + 4000x3 + 10 000x4 + (1 − 40x + 600x2 − 4000x3 + 10 000x4) = 2 + 1200x2 + 20 000x4
b (2 + 13 x)3 = 23 + 3(22)( 1
3 x) + 3(2)( 13 x)2 + ( 1
3 x)3 = 8 + 4x + 2
3 x2 + 127 x3
∴ (2 − 13 x)3 − (2 + 1
3 x)3 = 8 − 4x + 2
3 x2 − 127 x3 − (8 + 4x + 2
3 x2 + 127 x3)
= −8x − 227 x3
c = (1 + 4y)(1 + 3y + 3y2 + y3) = 1 + 3y + 3y2 + y3 + 4y + 12y2 + 12y3 + 4y4 = 1 + 7y + 15y2 + 13y3 + 4y4
d = (1 − x)(1 + 3x
+ 23x
+ 31x
)
= 1 + 3x
+ 23x
+ 31x
− x − 3 − 3x
− 21x
= −x − 2 + 22x
+ 31x
C2 SEQUENCES AND SERIES Answers - Worksheet D page 3
Solomon Press
8 a = (1 + x2)[1 + 10(−3x) + 10 9
2× (−3x)2 + 10 9 8
3 2× ×× (−3x)3 + …]
= (1 + x2)[1 − 30x + 405x2 − 3240x3 + …] = 1 − 30x + 405x2 − 3240x3 + x2 − 30x3 + … = 1 − 30x + 406x2 − 3270x3 + …
b = (1 − 2x)[1 + 8x + 8 72× x2 + 8 7 6
3 2× ×
× x3 + …] = (1 − 2x)[1 + 8x + 28x2 + 56x3 + …] = 1 + 8x + 28x2 + 56x3 − 2x − 16x2 − 56x3 + … = 1 + 6x + 12x2 + …
c = (1 + x + x2)[1 + 6(−x) + 6 52× (−x)2 + 6 5 4
3 2× ×
× (−x)3 + …] = (1 + x + x2)[1 − 6x + 15x2 − 20x3 + …] = 1 − 6x + 15x2 − 20x3 + x − 6x2 + 15x3 + x2 − 6x3 + … = 1 − 5x + 10x2 − 11x3 + …
d = (1 + 3x − x2)[1 + 7(2x) + 7 62× (2x)2 + 7 6 5
3 2× ×× (2x)3 + …]
= (1 + 3x − x2)[1 + 14x + 84x2 + 280x3 + …] = 1 + 14x + 84x2 + 280x3 + 3x + 42x2 + 252x3 − x2 − 14x3 + … = 1 + 17x + 125x2 + 518x3 + …
9 a = 84
× y4 × ( 1y
)4 = 70 b = 126
× (2y)6 × (− 12y
)6 = 924
c = 62
× ( 1y
)4 × (y2)2 = 15 d = 93
× (3y)6 × (− 21y
)3 = −61 236
10 a ( 1)2
n n− × ( 25 )2 = 1.6
n(n − 1) = 252 × 1.6 = 20
n2 − n − 20 = 0 (n + 4)(n − 5) = 0 n > 0 ∴ n = 5 b = 5 × ( 2
5 )4 = 16125 or 0.128
11 a y1 = (1 − 2x)[1 + 10x + 10 9
2× x2 + …]
= 1 + 10x + 45x2 − 2x − 20x2 + … = 1 + 8x + 25x2 + … ∴ a = 25, b = 8, c = 1 b x = 0.2: y1 = 0.6 × (1.2)10 = 3.71504 y2 = (25×0.04) + (8×0.2) + 1 = 3.6 % error =
3.71504 3.63.71504
− × 100% = 3.1% (2sf)
12 a (1 + px)q = 1 + q(px) + ( 1)2
q q− (px)2 + … ∴ pq = −12 and 1
2 p2q(q − 1) = 60
sub. p = − 12q
⇒ 72q
(q − 1) = 60
72(q − 1) = 60q q = 6, p = −2 b = 6 5 4
3 2× ×× × (−2)3 = −160
C2 SEQUENCES AND SERIES Answers - Worksheet D page 4
Solomon Press
13 a = 312 + 12(311)(−
3x ) + 12 11
2× (310)(−
3x )2 + 12 11 10
3 2× ×
× (39)(−3x )3 + …
= 531 441 − 708 588x + 433 026x2 − 160 380x3 + … b let
3x = 0.002 ∴ x = 0.006
2.99812 ≈ 531 441 − 4251.528 + 15.588 936 − 0.034 642 080 = 527 205.03 (2dp) 14 a = 1 − 5x + 10x2 − 10x3 + 5x4 − x5 b = 3 − 2 3 + 3 − 2 = 1 − 3 c i = [( 3 + 1)( 3 − 2)]5 = (1 − 3 )5 = 1 − 5( 3 ) + 10( 3 )2 − 10( 3 )3 + 5( 3 )4 − ( 3 )5 = 1 − 5 3 + 30 − 30 3 + 45 − 9 3 = 76 − 44 3 ii = ( 3 + 1)(76 − 44 3 ) = 76 3 − 132 + 76 − 44 3 = −56 + 32 3 15 a = 1 + 9(
2x ) + 9 8
2× (
2x )2 + 9 8 7
3 2× ×× (
2x )3 + 9 8 7 6
4 3 2× × ×
× × (2x )4 + …
= 1 + 92 x + 9x2 + 21
2 x3 + 638 x4 + …
b = 212 − (− 21
2 ) = 21 c = (1 × 63
8 ) + (2 × 212 ) = 7
828 16 10(x3)2( 2
ax
)3 = −80
a3 = −8 a = −2 17 a (1 + x
k)n = 1 + n( x
k) + ( 1)
2n n− ( x
k)2 + ( 1)( 2)
3 2n n n− −
× ( xk
)3 + …
∴ 2( 1)2
n nk− = 3 × 3
( 1)( 2)6
n n nk
− −
kn(n − 1) = n(n − 1)(n − 2) n(n − 1)[k − (n − 2)] = 0 n > 1 ∴ k − (n − 2) = 0 k = n − 2 b k = 7 − 2 = 5 (1 +
5x )7 = 1 + 7(
5x ) + 7 6
2× (
5x )2 + 7 6 5
3 2× ×× (
5x )3 + 7 6 5 4
4 3 2× × ×
× × (5x )4 + …
= 1 + 75 x + 21
25 x2 + 725 x3 + 7
125 x4 + …
Solomon Press
SEQUENCES AND SERIES C2 Answers - Worksheet E
1 = 1 + 4(4x) + 6(4x)2 + 4(4x)3 + (4x)4 2 a u5 = 3 × (−2)4 = 48
= 1 + 16x + 96x2 + 256x3 + 256x4 b S10 = 103[1 ( 2) ]
1 ( 2)− −− −
= −1023
c positive terms form GP: a = 3, r = (−2)2 = 4
S8 = 83(4 1)
4 1−
− = 65 535
3 a = 1 + 7(3x) + 7 6
2× (3x)2 4 GP: a = 8, r = 2, n = 10
+ 7 6 53 2× ×× (3x)3 + 7 6 5 4
4 3 2× × ×
× × (3x)4 + … S10 = 108(2 1)
2 1−
− = 8184
= 1 + 21x + 189x2 + 945x3 + 2835x4 + … b let x = 0.01 1.037 ≈ 1 + 0.21 + 0.0189 + 0.000 945 + 0.000 028 35 = 1.229 87 (5dp) 5 a = 25 + 5(24)x + 10(23)x2 6 a amount in account after 3rd payment in + 10(22)x3 + 5(2)x4 + x5 = 200 + (1.005 × 200) + (1.0052 × 200) = 32 + 80x + 80x2 + 40x3 + 10x4 + x5 = 603.005 b = 32 − 80x + 80x2 − 40x3 + 10x4 − x5 interest paid at end of 3rd month c (2 + 5 )5 = 32 + 80( 5 ) + 80( 5 )2 = 0.005×603.005 = £3.02 (nearest penny) + 40( 5 )3 + 10( 5 )4 + ( 5 )5 b amount paid in = 12 × 200 = £2400 = 32 + 80 5 + 400 + 200 5 + 250 + 25 5 amount in account after 12 months = 682 + 305 5 = 200(1.005 + 1.0052 + … + 1.00512) ∴ (2 + 5 )5 − (2 − 5 )5 = 200 × S12 [GP: a = 1.005, r = 1.005]
= (682 + 305 5 ) − (682 − 305 5 ) = 200 × 121.005(1.005 1)
1.005 1−
− = 2479.45
= 610 5 , k = 610 total interest = 2479.45 − 2400 = £79.45
7 = 1 + 8(−3x) + 8 7
2× (−3x)2 8 a Sn = a + ar + ar2 + … + arn − 1
+ 8 7 63 2× ×
× (−3x)3 + … rSn = ar + ar2 + … + arn − 1 + arn = 1 − 24x + 252x2 − 1512x3 + … subtracting, Sn − rSn = a − arn Sn(1 − r) = a(1 − rn)
Sn = (1 )1
na rr
−−
b r = 6 ÷ 3 = 2 a × 23 = 3 ∴ a = 3
8
S16 = 163
8 (2 1)2 1
−−
= 5824 575
C2 SEQUENCES AND SERIES Answers - Worksheet E page 2
Solomon Press
9 a = 1 + n(ax) + ( 1)2
n n− (ax)2 + … 10 = 26 + 6(25)(5x) + 6 52× (24)(5x)2 + …
= 1 + anx + 12 a2n(n − 1)x2 + … = 64 + 960x + 6000x2 + …
b 12 a2n(n − 1) = 3an
a2n(n − 1) = 6an an[a(n − 1) − 6] = 0 n ≠ 0 ∴ a(n − 1) − 6 = 0 an − a = 6 n = 6 a
a+
c n = 10 ∴ coeff. of x3 = 10 9 83 2× ×× × ( 2
3 )3 = 5935
11 a 162
1 r− = 486 12 a = 1 + 4(3x) + 6(3x)2 + 4(3x)3 + (3x)4
1 − r = 162486 = 1
3 ∴ r = 23 = 1 + 12x + 54x2 + 108x3 + 81x4
b u6 = 162 × ( 23 )5 = 64
3 or 1321 b term in x2 = (1)(54x2) + (4x)(12x) + (−x2)(1)
c S10 = 102
323
162[1 ( ) ]
1
−
− = 477.572 coefficient of x2 = 54 + 48 − 1 = 101
13 a time = 120 × (0.9)3 = 87.48 seconds 14 = [1+8(2x )+ 8 7
2× (
2x )2+…][1+6(−x)+ 6 5
2× (−x)2+…]
b GP: a = 120, r = 0.9, n = 12 = [1 + 4x + 7x2 + …][1 − 6x + 15x2 + …]
S12 = 12120[1 (0.9) ]
1 0.9−−
= 1 − 6x + 15x2 + 4x − 24x2 + 7x2 + …
= 861.08 seconds = 1 − 2x − 2x2 + … = 14 mins 21 secs (nearest sec.) ∴ A = −2, B = −2 15 a 6, 12, 24, 48 16 a = 1 + 4x + 6x2 + 4x3 + x4 b GP: a = 6, r = 2, n = 10 b = 1 − 4x + 6x2 − 4x3 + x4
S10 = 106(2 1)
2 1−
− = 6138 c (1 + 4x + 6x2 + 4x3 + x4)
+ (1 − 4x + 6x2 − 4x3 + x4) = 82 2 + 12x2 + 2x4 = 82 x4 + 6x2 − 40 = 0 (x2 + 10)(x2 − 4) = 0 x2 = −10 [no real solutions] or x2 = 4 x = ± 2
17 a a × (1.5)2 = 18 18 (1 + 2ax )10 + (1 + bx)10
a = 18 ÷ 2.25 = 8 = 1 + 10(2ax ) + 10 9
2× (
2ax )2 + …
b S6 = 68[(1.5) 1]
1.5 1−
− = 166.25 + 1 + 10(bx) + 10 9
2× (bx)2 + …
c 8 × (1.5)k − 1 > 8000 = 2 + (5a + 10b)x + ( 454 a2 + 45b2)x2 + …
(k − 1) lg 1.5 > lg 1000 ∴ 5a + 10b = 0 ⇒ a = −2b k > lg1000
lg1.5 + 1 and 45
4 a2 + 45b2 = 90 ⇒ a2 + 4b2 = 8
k > 18.04 ∴ smallest k = 19 sub. (−2b)2 + 4b2 = 8 b2 = 1 a < b ∴ b = 1, a = −2
Solomon Press
SEQUENCES AND SERIES C2 Answers - Worksheet F
1 a a = 108, ar3 = 32 2 = 1 + 5(−2x) + 10(−2x)2 ∴ r3 = 32 ÷ 108 = 8
27 + 10(−2x)3 + 5(−2x)4 + (−2x)5
r = 8327 = 2
3 = 1 − 10x + 40x2 − 80x3 + 80x4 − 32x5
u3 = 108 × ( 23 )2 = 48
b S∞ = 23
1081−
= 324
3 a new subscribers in 4th week 4 a = 1 + 7(4x) + 7 6
2× (4x)2 + …
= 200 × (1.15)3 = 304.175 = 1 + 28x + 336x2 + … = 304 (nearest unit) b (1 + 2x)2(1 + 4x)7 b new subscribers: GP, a = 200, r = 1.15 = (1 + 4x + 4x2)(1 + 28x + 336x2 + …)
S10 = 10200[(1.15) 1]
1.15 1−
− = 4060.74 term in x2
total no. of subscribers = 3600 + S10 = (1)(336x2) + (4x)(28x) + (4x2)(1) = 7661 (nearest unit) coefficient of x2 = 336 + 112 + 4 = 452 5 a = 1 + 2n( x
k) + 2 (2 1)
2n n− ( x
k)2 6 a r = 3 2 ÷ 6 = 3
+ 2 (2 1)(2 2)3 2
n n n− −× ( x
k)3 + … a = 6 ÷ 3 = 2
= 1 + 2nk
x + 2(2 1)n n
k− x2
+ 32 ( 1)(2 1)
3n n n
k− − x3
+… b S8 = 82[( 3) 1]
3 1−
−
b 32 ( 1)(2 1)
3n n n
k− − = 1
2 × 2
(2 1)n nk
− = 80 2 3 13 1 3 1
+×− +
4n(n − 1)(2n − 1) = 3kn(2n − 1) = 80 2( 3 1)3 1
+−
n(2n − 1)[4(n − 1) − 3k] = 0 = 40 2( 3 1)+ n > 1 ∴ 4(n − 1) − 3k = 0 3k = 4(n − 1) c 2n
k = 2 ∴ n = k
∴ 3k = 4k − 4 k = 4, n = 4
7 9
13r
r =∑ : GP, a = 3, r = 3 8 a = 1 + 9(2x) + 9 8
2× (2x)2 + 9 8 7
3 2× ×× (2x)3 + …
S9 = 93(3 1)
3 1−
− = 29523 = 1 + 18x + 144x2 + 672x3 + …
∴ 9
1(3 1)r
r=−∑ = 29523 − 9 b (1 − 2x)9 = 1 − 18x + 144x2 − 672x3 + …
= 29 514 ∴∴∴∴ (1 + 2x)9 + (1 − 2x)9 = (1 + 18x + 144x2 + 672x3 + …) + (1 − 18x + 144x2 − 672x3 + …) = 2 + 288x2 (ignoring terms in x4 and higher)
c let x = 0.001 ∴∴∴∴ 1.0029 + 0.9989 ≈ 2 + 0.000 288 = 2.000 288 (7sf)
C2 SEQUENCES AND SERIES Answers - Worksheet F page 2
Solomon Press
9 (k − x)9 = k9 + 9(k8)(−x) + 9 8
2× (k7)(−x)2 + … 10 = 34 + 4(3)3(2x) + 6(3)2(2x)2
= k9 − 9k8x + 36k7x2 + … + 4(3)(2x)3 + (2x)4 ∴ −b = −9k8 and b = 36k7 = 81 + 216x + 216x2 + 96x3 + 16x4 9k8 = 36k7 9k7(k − 4) = 0 k ≠ 0 ∴ k = 4 a = k9 = 262 144 b = 9k8 = 589 824 11 a
1t
r− = 3t 12 a = 1 + 4(−2x) + 6(−2x)2 + 4(−2x)3 + (−2x)4
1 − r = 3tt
= 13 ∴ r = 2
3 = 1 − 8x + 24x2 − 32x3 + 16x4
b 42
323
[1 ( ) ]
1
t −
− = 130 b let x = y2 − 2y
t = ( 13 × 80) ÷ 65
81 = 54 (1 + 4y − 2y2)4 = 1 − 8(y2 − 2y) + 24(y2 − 2y)2 + … term in y2 = −8y2 + 24(−2y)2 coefficient of y2 = −8 + 96 = 88 13 a = 12000 × (0.75)4 14 a p(−2) = 14 − (−1)4 = 1 − 1 = 0 = 3796.875 ∴ (x + 2) is a factor of p(x) = £3800 (3sf) b p(x) = [x4 + 4(x3)(3) + 6(x2)(32) + 4(x)(33) + 34] b GP: a = 12000, r = 0.75 − [x4 + 4x3 + 6x2 + 4x + 1]
S8 = 812000[1 (0.75) ]
1 0.75−
− = 8x3 + 48x2 + 104x + 80
= £43 200 (3sf) = 8(x3 + 6x2 + 13x + 10) p(x) = 8(x + 2)(x2 + 4x + 5) c 8(x + 2)(x2 + 4x + 5) = 0 x = −2 or (x2 + 4x + 5) = 0 b2 − 4ac = 16 − 20 = −4 b2 − 4ac < 0 ∴ no real sols to (x2 + 4x + 5) = 0 ∴ only one real solution to p(x) = 0
x2 + 4x + 5 x + 2 x3 + 6x2 + 13x + 10
x3 + 2x2 4x2 + 13x 4x2 + 8x 5x + 10 5x + 10
C2 SEQUENCES AND SERIES Answers - Worksheet F page 3
Solomon Press
15 a (1 − x)(1 + 2x)n 16 = ( 3
x)4 + 4( 3
x)3(−x) + 6( 3
x)2(−x)2
= (1 − x)[1 + n(2x) + ( 1)2
n n− (2x)2 + …] + 4( 3x
)(−x)3 + (−x)4
= (1 − x)[1 + 2nx + 2n(n − 1)x2 + …] = x4 − 12x2 + 54 − 2108x
+ 481x
∴ 2n(n − 1) − 2n = 198 n2 − 2n − 99 = 0 (n + 9)(n − 11) = 0 n ≥ 0 ∴ n = 11 b (1 − x)(1 + 2x)11 = (1 − x)[… + 11 10
2× (2x)2 + 11 10 9
3 2× ×
× (2x)3 + …] = (1 − x)[… + 220x2 + 1320x3 + …] ∴ coefficient of x3 = 1320 − 220 = 1100 17 a S4 = 34 − 1 = 80 18 a 3(x − 3) = y − 3 S3 = 33 − 1 = 26 y = 3x − 6
u4 = S4 − S3 = 80 − 26 = 54 b 3
3x
=
3y
b Sn − 1 = 3n − 1 − 1 x3 = 9y = 9(3x − 6) un = Sn − Sn − 1 x3 − 27x + 54 = 0 = (3n − 1) − (3n − 1 − 1) c trying x = 1, 2 etc. ⇒ x = 3 is a solution = 3n − 3n − 1 ∴ (x − 3) is a factor = 3n(1 − 1
3 ) = 23 (3n) [k = 2
3 ] c un − 1 = 2
3 (3n − 1) un ÷ un − 1 = 2
3 (3n) ÷ 23 (3n − 1) = 3
un ÷ un − 1 is constant ∴ geometric (x − 3)(x2 + 3x − 18) = 0 (x − 3)(x + 6)(x − 3) = 0 x = −6 or 3
x2 + 3x − 18 x − 3 x3 + 0x2 − 27x + 54
x3 − 3x2 3x2 − 27x 3x2 − 9x − 18x + 54 − 18x + 54
