cape pure mathematics unit 2module 2: sequences, series and approximations

7/27/2019 CAPE PURE MATHEMATICS UNIT 2MODULE 2: SEQUENCES, SERIES AND APPROXIMATIONS

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CAPE Pure Mathematics Unit 2

Practice Questions

By Carlon R. Baird

MODULE 2: SEQUENCES, SERIES AND APPROXIMATIONS

1. (a) Show that 1 ( 1) ( 1)2

r r r r r .

(b) Hence show using method of differences that 1

12

n

r

nr n

.

(c) Evaluate20

10

4r

r

.

2. (a) Given that 1 1( 1)! ! ( 1)!

r

r r r

find

1 ( 1)!

n

r

r

r

(b)1

( ) ,( 1)

f p pp p

+

(i) Show that ( ) ( 1)( 1)( 2)

vf p f p

p p p

, stating the value ofv.

(ii) Hence show that by method of differences, that

2

1

1 (2 3)

( 1)( 2) 4( 1)(2 1)

n

p

n nS

p p p n n

(iii) Deduce the sum to infinity ofS.

3. (a) Prove by the method of mathematical induction, that, forn +,1

2 2 1 ( 1)2n

r n

r

r n

(b) Prove by induction that forn +, that 1

1(3 4) 3 11 .

2

n

r

r n n


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4. (a) The expressions 26, 2 , andx x x form the first three terms of ageometric progression. By calculating two different expressions for

the common ratio, form and solve an equation inx to find possible

values of the first term.

(b) Dylan invest $D at a rate of interest 4% per annum. After 5 years it

will be worth $10,000. How much (to the nearest penny) will it be

worth after 10 years.

(c) The first three terms of a geometric series are (3 1), (2 2) andt u t u

(2 1)t u where tand u are constants.

(i)

Use an algebraic method to show that one possible value ofu is5 and to find the other possible value ofu.

(ii) For each possible value ofu, calculate the value of the commonratio of the series.

Given that 5u and that the sum to infinity of the geometric series is896, calculate:

(iii) The value oft.(iv) The sum of the first twelve terms of the series giving answer to

2 decimal places.

5. (a) For the arithmetic series 5 9 13 17 ... Find:

(i) The 20th term(ii) The sum of the first 20 terms.

(b) The sum of the first two terms of an arithmetic series is 47.

The thirtieth term of this series is 62 .

Find:

(i) The first term of the series and the common difference


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(ii) The sum of the first 60 terms of the series.6. (a) Find the first four terms of the of the sequence:

1 14, 7

n nu u u

(b) A sequence of terms { nU }, 1n is defined by the recurrence relation

2 1where is a constant

n n nU U U

Given also that1 2

2 and 5U U :

(i) Find an expression in terms offor3

U

(ii) Find an expression in terms of for 4U Given that the value of

421U :

(iii) Find the possible values of(c) Given that

4 3

4 2

10 1r

r ry

r r

where 1r . Show that

ry is

convergent. Hence state the limit it converges to.

7. A sequence1 2 3 4, , , ,...u u u u is defined by 1 15 3(2 ), 7

n

n nu u u

(a) Determine the first four terms of the sequence.(b)Prove by mathematical induction forn +, that 5 2n n

nu .

8. (a) Use Maclaurins theorem to find the first three non-zero terms in theseries expansion of

(1 2 )ln

1 3

x

x

, and state the interval inx for

which the expansion is valid.

(b) (i) Show using Maclaurins theorem that2

2 33( 3)sin3 3 3 ...2

x

e x x x x

where is a constant.


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(ii) Given that the first non-zero term in the expansion, in

ascending powers ofx, of 3sin3 ln(1 ) is ,xe x x x x

where is a constant, find the values of , and .

9. (a) Show that the Taylor expansion ofsin( )x in ascending powers of6

x

up to the term

2

6x

is

2

1 3 1sin( )

2 2 6 4 6x x x

.

(b) Using the series in (a) find, in terms of , an approximation for

2sin

9

.

10. Given that3

cos( ) sin( ) 2 0

dy

x y x ydx and that 1y at 0x , use

Taylors method to show that, close to 0x , terms in 4x and higher powers

can be ignored,2 311 561 2

2 3y x x x .

11. (a) Expand fully the expression 3(1 3 )(1 2 ) .x x (b) Expand 3(2 )y . Hence or otherwise, write down the expansion

2 3(2 )x x in ascending powers ofx.

(c) The coefficient of 2x in the expansion of 3(2 )(3 )x bx is 45. Findthe possible values of the constant b.

(d) Find the term independent ofx in the expansion of 32 1 .2

xx

12. (a) Use the binomial series to expand 102 3x in ascending powers ofxup to and including the term in

3x , giving each coefficient as an

integer.

(b) Use your series expansion, with suitable value forx, to obtain anestimate for 1.9710, giving your answer to 2 decimal places.


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13. (a) Find the binomial expansion of2

1

x

x

in ascending powers ofx as

far as the term in3

x . State the range of values ofx for which the

expansion is valid.

(b) Find an expansion of (1 2 )x up to and including the term in 3x . Bysubstituting in 0.01x , find a suitable decimal approximation to 2

(c) (i) Express 26 7 5(1 )(1 )(2 )

as partial fractions.

(ii) Hence or otherwise expand 26 7 5(1 )(1 )(2 )

in ascending

powers ofas far as the term in3 .

(iii) State the set of values offor which the expansion is valid.14. (a) Evaluate 9!

2!3!4!.

(b) Prove that2! 2( 1)! ( 1)! ( 2 )

( )! ! ( 1)!( 1)! ( 1)! !

n n nn nr n r

n r r r n r n r r

(c) Prove that n nr n rC C 15. (a) ( ) 2 3xf x x

(i) Show that there exist a root in the interval [2, 3] using theintermediate value theorem.

(ii) Using the end points of this interval by interval bisection,obtain a first and second approximation tox.


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(b) (i) Using the intermediate value theorem show that one root of theequation

3 7 2 0x x lies in the interval [2, 3].

(ii) Use interval bisection to find the root to two decimal places.16. (a) Show that a root of the equation 2 cos 1 0x x lies in the interval

[1, 1.5].

(b) Find this root using linear interpolation correct to two decimal places.17. 3 2( ) 3 5 4f x x x x

Taking 1.4 as a first approximation to a root,x, of this equation, use

Newton-Raphson process once to obtain a second approximation tox. Give

your answer to three decimal places.


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By Carlon R. Baird


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1. (a) R.T.S : 1 ( 1) ( 1)2

r r r r r

R.H.S: 2 21 1( 1) ( 1)2 2

r r r r r r r r

1 22

r

r

(b) By method of differences:

1 1

1( 1) ( 1)

2

n n

r r

r r r r r

(c)

Recall that :

1

1 1( ) ( ) ( )

n n k

r k r r f r f r f r

20 20

10 10

4 4r r

r r

1

1 1

1( 1) ( 1)2

1( 1) ( 1)

2

11(2) 2(3) 3(4) ... ( 1)( 1 1) ( 1)

2

1(1 1) 2(1) 3(2) 4(3) ... ( 1)

122

n

r

n n

r r

r r r r

r r r r

n n n n

n n

6 12 ... ( 1)n n ( 1)0 2

n n

6 12 ... ( 1)n n

1( 1)

2

12

n n

nn


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20 9

1 1

4

20 94 20 1 9 1

2 2

4 10(21) 9(5)

4 210 45

660

r r

r r

2. (a) Given that 1 1( 1)! ! ( 1)!

r

r r r

(b) 1( ) ,( 1)

f p pp p

+

(i) R.T.S: ( ) ( 1)( 1)( 2)

vf p f p

p p p

L.H.S:

1 1

1 1

1 1

1 ! ! ( 1)!

1 1

! ( 1)!

1 1 1 1 1 1 1 1 11 ... ...

2! 3! 4! ! 2! 3! 4! ( 1 1)! ( 1)!

11

2!

n n

r r

n n

r r

r

r r r

r r

n n n

1

3!

1

4! ...

1

!n

1

2!

1

3!

1

4! ...

1

!n

1

( 1)!

11

( 1)!

n

n

1 1( ) ( 1)

( 1) ( 1)( 1 1)

f p f pp p p p


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1 1

( 1) ( 1)( 2)

( 2)

( 1)( 2)

2( 1)( 2)

p p p p

p p

p p p

p p p

2v

(ii)2

1

1 (2 3)R.T.S :

( 1)( 2) 4( 1)(2 1)

n

p

n n

p p p n n

2 2

1 1

2 2 2

1 1 1

1 1 2

( 1)( 2) 2 ( 1)( 2)

1 2 1 1 1

2 ( 1)( 2) 2 ( 1) ( 1)( 2)

1 1 1 1 1 1...

2 1(2) 2(3) 3(4) 4(5) 2 (2 1)

n n

p p

n n n

p p p

p p p p p p

p p p p p p p

n n

1 1 1 1...

2(3) 3(4) 4(5) (2 1 1)(2 1 2)1

+(2 1)(2 2)

1 1 1

2 2 6

n n

n n

1

12

1

20 ...

1

2 (2 1)n n

1

6

1

12

1

20

1+

2 (2 1)n n

2

1

(2 1)(2 2)

1 1 1

2 2 (2 1)(2 2)

1 (2 1)(2 2) 2

2 2(2 1)(2 2)

1 4 4 2

2

n n

n n

n n

n n

n n n

2 2

2 2( 1)(2 1)n n


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2

2

1 4 6

2 4( 1)(2 1)

1 2(2 3 )

2 4( 1)(2 1)

n n

n n

n n

n n

2

1

1 (2 3)

( 1)( 2) 4( 1)(2 1)

n

r

n nS

r r r n n

(iii) (2 3)lim lim4( 1)(2 1)n n

n nS

n n

2

2

2

2

2

2 2

2

2 2 2

2

2 3lim4 2 2 1

2 3lim

8 12 4

2 3

lim8 12 4

32

lim12 4

8

2 0

8 0 0

1

4

n

n

n

n

n nn n n

n n

n n

n n

n nn n

n n n

n

n n


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3. (a) Let Pnbe the statement

1

2 2 1 ( 1)2n

r n

r

r n

Showing

1P is true:

L.H.S.:1

1

1

2 1(2) 2r

r

r

R.H.S.: 12 1(1 1)2 2 1 2(0)

2

1

L.H.S R.H.S

P is true

Assume Pkis true:

1

2 2 1 ( 1)2k

r k

r

r k

Verifying1

Pk

is true

1

1

1

1

1

1 1

1

1

1

1

1

P P ( 1) 2

2 1 ( 1)2 ( 1) 2

2 2( 1)2 ( 1) 2

2 2 2 ( 1) 2 ( 1)

2 2 ( 1) 2 ( 1)

2 2 ( 1) ( 1)

2 2 2 1 1

2 2 2( 1) 2

2 2 2 ( 1) 1

2 1 ( 1) 1 2

k

k k

k k

k k

k k

k k

k

k

k

k

k

k

k k

k k

k k

k k

k k

k

k

k

k

1P is true

k

By Principle of Mathematical Induction Pnholds true n +


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(b) Let Pnbe the statement

1

13 4 3 11

2

n

r

r n n

Showing

1P is true:

L.H.S:

1

1(3 4) 3(1) 4 7

rr

R.H.S: 1

1 3(1) 11 72

1

L.H.S R.H.S

P is true

Assume Pkis true:

1

13 4 3 11

2

k

r

r k k

Verifying

1P

kis true:

1

2

2

2

P P 3( 1) 4

13 11 3 3 4

2

13 11 2 3 7

2

13 11 6 14

2

13 17 14

2

13 3 14 14

2

13 ( 1) 14( 1)

2

11 3 14

2

11 3 3 11

2

11 3( 1) 11

2

k kk

k k k

k k k

k k k

k

k k k

k k k

k k

k k

k k


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1

P is truek

By Principle of Mathematical Induction Pnholds true n +

4. (a)

From equn :2x

ra

From equn :2

2 xra

Substituting rinto equn

2 2

2 2

2

2 2

2

4

4

x x

a a

x x

a a

x ax

From equn 6a x 2 2

2 3 2

3 2

2

4 ( 6)

4 6

10 0

( 10) 0

0 or 10

x x x

x x x

x x

x x

x x

Possible values of the first term:

0 6 6 6

or

10 6 4 4

a a

a a

2 2

6

2

a x

ar x

ar x


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(b) $a D After 1 year:

4$

100ar D D

4$ 100

100 4$

100

104$

100

DD

D D

D

$1.04

1.04 1.041.04

ar D

D Dr

a D

After 2 years:2 2$(1.04)ar D

Given that after 5 years it will be worth $10,0005 5

5

(1.04) $10,000

10000$ $8219.27

(1.04)

ar D

D

So Dylans initial investment was about $8219.27

Now, after 10 years, i.e 10ar ,

1010

5

100001.04

(1.04)

12166.52902

ar

The investment will be worth $12166.53

(c) (3 1)a t u 2

(2 2)(2 1)

ar t uar t u

(i) (2 2) 2 2(3 1) 3 1

t u uar a r

t u u

Rewriting another equation for the third term of the GP:


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2

2

2

2

2

(2 2)(3 1)

(3 1)

(2 2)(3 1)

(3 1)(2 2)

3 1

uar t u

u

ut u

ut u

u

Now we could say that:2(2 2)

(2 1)3 1

t ut u

u

2

2 2

2

2

(2 1)(3 1) (2 2)

6 2 3 1 4 8 4

2 9 5 0

2 10 5 0

2 ( 5) 1( 5) 0

( 5)(2 1) 0

5

or

u u u

u u u u u

u u

u u u

u u u

u u

u

u

1

2

(ii) When 5u ; 2(5) 2 12 33(5) 1 16 4

r

When1

;2

u

12 2

1 222

3113 1

22

r


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(iii) Given that 5u and 896S (3(5) 1)

89631

14

16896

1

4

16 224

14

a tS

r

t

t

t

(iv) 11

n

n

a rS

r

12

12

3224 1

4

31

4

216.9044971...=

1

4

=867.61798...=867.62 {2 d.p.}

S

5. (a) 5+9+13+17+...(i) 5a

9 5 4d

( 1)nu a n d

20

5 (20 1)(4)

5 (19)(4)

81

u


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(ii) 2 ( 1)2

n

nS a n d

20

202(5) (20 1)(4)

2

10 10 76

=860

S

(b) (i) 2 ( 1)2

n

nS a n d

2

2

22 (2 1)

2

2 47

S a d

S a d

30

( 1)

29 62

nu a n d

u a d

We have two simultaneous equns:2 47a d

29 62a d

Equn 62 12a d

Substituting a into equn

2( 62 29 ) 47

124 58 47

57 171

3

d d

d d

d

d

26a

(ii)

60

602( 26) 59( 3)

2

30 52 177

6870

S


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6. (a)

The first four terms of the sequence:

7,11,15,19,...

(b)2 1

,n n n

U U U 1 22 and 5U U

(i)

(ii)

(iii) Given that4

21U 2

2

2

5 2 5 21

5 2 16 0

5 10 8 16 0

5 ( 2) 8( 2) 0

(5 8)( 2)=0

8= or = 25

(c) 4 34 2

10 1r

r ry

r r

, where 1r

1

1

2 1 1 1

3 2

4 3

4

7

4 7 4 11

4 11 4 154 15 4 19

n nu u

u

u u u

u u

u u

3 1 2 1 1 1

2 1

5 2

U U U U U U

4 3 2

2

(5 2) 5

5 2 5

U U U


20/39

4 3

4 2

4 3

4 4 4

4 2

4 4

4

2

10 1lim lim

10 1

lim

1 110

lim1

1

10 0 0

1 0

10

rr r

r

r

r ry

r r

r r

r r rr r

r r

r r

r

As lim 10, is convergent

i.e it converges to the limit 10

r rr

y y

7. (a)

(b) Let Pnbe the statement 5 2

n n

nu

Showing1

P is true:

1

1

1

2 1

2

3 2

3

4 3

5 3(2 )

7

5 3(2 )

5(7) 3(2)=29

5 3(2 )

5(29) 12

=133

5 3(2 )

5(133) 3(8)

641

n

n nu u

u

u u

u u

u u


21/39

1 1

1

1 1

1

5 2 7

P is true

u

u u

Assumek

P is true:

5 2k kk

u

Verifyingk+1

P is true:

1

1

1

1 1

5 3(2 )

=5 5 2 3(2 )

5 5 5 2 3 2

5 2 (5 3)

5 2 (2)

5 2

k

k k

k k k

k k k

k k

k k

k k

u u

1P is true

k

By Principle of Mathematical Induction P holds truen

n +

8. (a) Let (1 2 )( ) ln1 3

xh x

x

1

2

1 1

( ) ln(1 2 ) ln(1 3 )

1 2 3'( )

2 1 2 1 3

1 3

1 2 1 3

(1 2 ) 3(1 3 )

h x x x

h xx x

x x

x x

2 2

2 2''( ) 1(2)(1 2 ) 3( 3)(1 3 )

2(1 2 ) 9(1 3 )h x x x

x x

3 3

3 3

'''( ) 4(2)(1 2 ) 18( 3)(1 3 )

8(1 2 ) 54(1 3 )

h x x x

x x


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2 2

3 3

1(0) ln(1) ln(1) 0

2

1 3'(0) 1 3 4

1 2(0) 1 3(0)

''(0) 2(1 0) 9(1 0) 2 9 7

'''(0) 8(1 2(0)) 54(1 3(0)) 8(1) 54 62

h

h

h

h

By Maclaurin's theorem:

2 3''(0) '''(0)( ) (0) '(0) ...2! 3!

h hh x h h x x x

2 3

2 3

7 62( ) 0 4 ...

2! 3!

(1 2 ) 7 31ln 4 ...

1 3 2 3

h x x x x

xx x x

x

where1 1

3 3x

(b) (i) Let ( ) sinxf x e x

'( ) 3cos3 sin3

3cos3 sin3

x x

x

f x e x x e

e x x

2

2

''( ) 9sin3 3 cos3 3cos3 sin3

9sin3 3 cos3 3 cos3 sin3

9 sin3 6 cos3

x x

x

x

f x e x x x x e

e x x x x

e x x

2 2

2 2 2

2 2 3

2 3

'''( ) 3( 9)cos3 18 sin3 ( 9)sin3 6 cos3

3( 9)cos3 18 sin3 ( 9)sin3 6 cos3

3 27 6 cos3 9 18 sin3

9 27 cos3 27 sin3

x x

x

x

x

f x e x x x x e

e x x x x

e x x

e x x

(0 )(0) sin(0) 0f e


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(0 )'(0) 3cos3(0) sin3(0) 3f e

(0) 2''(0) 9 sin3(0) 6 cos3(0) 6f e

(0) 2 3

2'''(0) 9 27 cos3(0) 27 sin3(0)

9 27f e

2 3


''(0) '''(0)( ) (0) '(0) ...

2! 3!

f ff x f f x x x

2

2 3

2 2 3

9 276( ) 0 3 ...

2! 3!

93 3 3 ...3!

f x x x x

x x x

22 33 3sin3 3 3 ...2

xe x x x x

(ii) Let ( ) ln(1 )q x x 1

'( ) (1 )1q x xx

2

2 2

''( ) ( )(1 )

(1 )

q x x

x

2 3

3 3

'''( ) 2( )( )(1 )

2 (1 )

q x x

x

1

2 2 2 2

3 3 3 3

(0) ln(1 (0)) ln(1) 0

(0) (1 (0)) (1)''(0) (1 (0)) (1)

'''(0) 2 (1 (0)) 2 (1) 2

q

qq

q


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2 3

2 3

2 3

2 2 3 3


''(0) '''(0)( ) (0) '(0) ...

2! 3!

( ) 2

0 ...2! 3!

1 1...

2 3

q qq x q q x x x

x x x

x x x

2 2 3 31 1ln(1+ )= ...2 3

x x x x

Hence,

In the question we were told that the first non-zero term in the

expansion of3sin3 ln(1 ) is ,xe x x x x , this means that the co-

efficient of bothx andx2 are 0.

2 02

2

2 3

2 2 3 3

2 2 2

2

3 3 3

2 3

2 2 3

2 3

2 2 3

3 3sin3 ln(1 ) 3 3 2

1 1...

2 3

13 3

2

3 3 1...

2 3

1 3 9(2 ) 3 ...2 2 2 3

1 3 9(2 ) 3 ...

2 2 3 2

xe x x x x x x

x x x x

x x x x x

x x

x x x

x x x


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2

2

2

13 0

2

13

2

1 43 ( 2) 22 2

2

3

2 3

2

3

3 9

2 3 2

23

2 93

2 3 2

2 8 9

3 3 2

13

2

9. (a) Let ( ) sinf x x | 1sin6 6 2

f

'( ) cosf x x |3

' cos6 6 2

f

''( ) sinf x x |1

'' sin6 6 2

f

Using Taylors expansion:

2''( )( ) ( ) '( ) ...

2!

f af x f a f a x a x a

2

2

11 3 2( ) ...2 2 6 2! 6

1 3 1sin( ) ...

2 2 6 4 6

f x x x

x x x


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(b)

10. (a) 3cos( ) sin( ) 2 0dyx y x ydx

Differentiating equn:

3

2

2

2

2

2

cos( ) sin( ) 2 0

cos( ) sin( ) cos( ) sin( ) 6 0

cos( ) sin( )

d dy d d

x y x ydx dx dx dx

d y dy dy dyx x y x x y

dx dx dx dx

d y dyx x

dx dx

cos( ) sin( )dy

y x xdx

2

2

2

2

6 0

cos( ) cos( ) 6 0

dyy

dx

d y dyx y x y

dx dx

Now, differentiating equn

2

2

2cos( ) cos( ) 6 0

d d y d d dyx y x y

dx dx dx dx dx

Given the initial conditions 0 01 at 0y x

2

2

2

2 1 3 2 1 2sin ...

9 2 2 9 6 4 9 6

1 3 1

2 2 18 4 181 3 1

2 36 1296

3 2 2

2

3 2 2

23 2 2

2

3 2 2

cos( ) sin( ) ( sin( )) cos( ) 6 12 0

cos( ) sin( ) sin( ) cos( ) 6 12 0

d y d y dy d y dy dyx x y x x y y

dx dx dx dx dx dx

d y d y dy d y dyx x y x x y y

dx dx dx dx dx


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3

0

0

0

cos(0) (1)sin(0) 2(1) 0

0 2 0

2

dy

dx

dy

dx

dy

dx

Substituting values of0

x ,0

0

anddy

ydx

into equn

2

2

2

0

2

2

0

2

2

0

cos(0) (1)cos(0) 6 1 2 0

0 1 0 12 0

11

d y

dx

d y

dxd y

dx

Substituting values of0

x ,2

0 2

0 0

, anddy d y

ydx dx

into equn

3

2 2

3

0

3

3

0

3

3

0

cos(0) sin(0) 11 (1)sin(0) cos(0) 2 6 1 11 12 1 2 0

0 0 2 66 48 0

112

d y

dx

d y

dx

d y

dx

To summarize:2 3

0 0 2 3

0 0 0

0, =1 , = 2 , 11, 112dy d y d y

x ydx dx dx

Now using Taylors expansion:

2 3

2 32 3

0 0

0 0 2 3

0 0 0

''( ) '''( )( ) ( ) '( ) ...2! 3!

...2! 3!

f a f af x f a f a x a x a x a

x x x xdy d y d yy y x x

dx dx dx


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29/39

3 2 1 2 33 3 3

1 2

2 2 3 3

3 3 3 3

27 27 9

bx C bx C bx bx

bx b x b x

3 2 2 3 3

2 2 3 3 2

2 3 2 27 27 9

54 54 18 2 3 27 ...

x bx x bx b x b x

bx b x b x x bx

Now considering the coefficients of 2x

2

2

2

2

18 27 45

18 27 45 0

out by 3

6 9 15 0

6 15 6 15 03 (2 5) 3(2 5) 0

2 5 3 3 0

5or 1

2

b b

b b

b b

b b bb b b

b b

b b

(d)

3The term independent of is

4x

12. (a)

10 10 0 9 1 8 2 7 310 10 10 10

0 1 2 32 3 2 3 2 3 2 3 2 3 ...x C x C x C x C x

2 31024 15360 10368 414770 ...x x x

(b)We first must find the value ofx obtaining an estimate for 101.97 2 3 1.97

3 2 1.97

0.01

x

x

x

3 0 1 2 3

3 2 1 02 3 2 3 2 3 2 3 2

0 1 2 3

6 4 4

2 3

6 3

3

1 1 1 1 1

2 2 2 2 2

1 1 1

3 32 4 8

3 3 1

2 4 8

x C x C x C x C xx x x x x

x x xx x x

x xx


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Now we can substitute x into our series expansion:

10 2 31.97 1024 15360(0.01) 103680(0.01) 414770(0.01) ...

1024 153.6 10.368 0.41477

880.35323

880.35 2 d.p.

13. (a) 1 12 22 2 2 11 1

x xx x

x x

Using the binomial expansion:

2( 1) ( 1)( 2)

1 1 ...2! 3!

n n n n n nx nx x

1

2

12 2 1

2

12 1

2

x x

x

12 3

2

2 3

2 3

1 1 1 1 11 1 2

1 1 1 2 2 2 2 22 1 2 1 ...

2 2 2 2! 2 3! 2

1 1 12 1 ...

4 8 4 16 8

1 1 12 1 ...

4 32 128

x xx x

x xx

x x x

where1

12

x

1

2 32

2 3

1 1 1 1 11 1 2

1 2 2 2 2 21 1 ...

2 2! 3!

1 3 51 ...

2 8 16

x x x x

x x x


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1 1

2 3 2 32 2

1 1 1 1 3 52 1 2 1 ... 1 ...

4 32 128 2 8 16x x x x x x x x

2 3 2 3 2 3 3

2 2 2 3 3 3 3

2 3

1 3 5 1 1 3 1 1 12 1 ...

2 8 16 4 8 32 32 64 128

1 1 3 1 1 5 3 1 12 1 ...

2 4 8 8 32 16 32 64 128

1 7 252 1 ...

4 32 128

x x x x x x x x x

x x x x x x x x x

x x x

Valid if 1 and 12

xx

1 for both to be validx

(a) 121 2 1 2x x

2 3

2 3

2 3

1 1 1 1 11 2 1 2 2

1 2 2 2 2 21 2 ...

2 2! 3!

1 11 4 8 ...

8 161 11 ...

2 2

x x

x

x x x

x x x

This expansionis valid for 2 1

1

x

x

Now substituting in 0.01x


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491 2(0.01) 0.98

50

49 149

50 50

17

2 25

7

5 2

2 37 1 1

1 0.01 0.01 0.01 ...2 25 2

1 0.01 0.00005 0.0000005 ...

0.9899495

7 0.989945 5 2

72

0.0989945 5

2 1.41421982

(b) (i) Let

26 7 5( )

1 1 2 1 1 2

A B CP

Multiplying both sides by 1 1 2


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2

2

2

6+7 5 1 2 1 2 1 1

Let 1

6 7 1 5 1 0 2 2 3 2 018 6

B=3

Let 1

6 7 1 5 1 (2)(1) (0)(1) (0)(2)

4=2

2

Let

A B C

A B CB

A B C

A

A

22

6 7 2 5 2 (3)(0) ( 1)(0) ( 1)(3)

12 3

4

A B C

C

C

2 3 4( )

1 1 2P

Valid 1 12

2

1 1 1

2 3 2 3 2 3

2 3 2 3 2 3

2 3

22 3

2 1 3 1 4 2

1 12 2 2 2 3 3 3 3 2 ...

2 4

1 12 2 2 2 3 3 3 3 2 ...

2 4

53 2 6 ...4

6 7 5 53 2 6 ...

1 1 2 4


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(ii) ss

(iii) All expansions valid for |x|


35/39

14. (a)

(b) 2! 2( 1)! ( 1)!

R.T.P : 2( )! ! ( 1)!( 1)! ( 1)! !

n n nn nr n r

n r r r n r n r r

N.B.

! ( 1)!

( 1)! ( 1)( )!

( 1)!( )!

( 1)! ( 1)!

r r r

n r n r n r

n rn r

n r

n n n

! 2( 1)!L.H.S:

( )! ! ( 1)!( 1)!

! 2( 1)!

( 1)! ( 1)!( 1)!( 1)!

( 1)

!( 1) 2( 1)!

( 1)!( 1)! ( 1)!( 1)!

!( 1) 2 ( 1)!

( 1)!( 1)!

( 1)!( 1) 2 ( 1)!

( 1)

n n

n r r r n r

n n

n r r n rr r

n r

n n r n

r n r r r n r

n n r r n

r n r r

n n n r r n

r n r

2

!( 1)!

( 1)! 1 2

1 ! 1 !

( 1)! 2( 1)! !

r

n n n r r

n r r r

n n nr n r n r r

9! 9 8 7 6 5 4!

2!3!4!

2 1 3 2 1 4!

15120

12

1260


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(c) R.T.P: n nr n rC C

i.e.

! !

!( )! ( )! !

n n

r n r n r n n r

! !

R.H.S:( )!( )!( )! !

!=

!( )!

n n

n r n n r n r n n r

n

r n r

15. (a) ( ) 2 3xf x x (i)

(ii)

To summarize: 1st approximation=2.5

2nd approximation=2.25

(b)

(i) Let

3

( ) 7 2f x x x

3

3

(2) 2 7(2) 2 4 0

(3) 3 7(3) 2 8 0

(2) (3) 0

By the I.V.T such that ( ) 0 in the interval 2,3

f

f

f f

f x

2

3

(2) 2 2 3 1 0

(3) 2 3 3 8 0

(2) (3) 0

By the Intermediate Value Theorem(I.V.T) such that ( ) 0

in the interval 2,3 .

f

f

f f

f x

2 3(2.5) 0.156854 0

2 2

2.0 2.5

2.0 2.5(2.25) 0.49317 0

2 2

2.25 2.5

a bf f f

a bf f f


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(ii)

2 3(2.5) 0.125 0

2 2

2<


38/39

16. (a) Let ( ) 2 cos( ) 1g x x x

(1) 2(1)cos(1) 1 0.0806046 0

(1.5) 2(1.5)cos(1.5) 1 0.78778... 0

(1) (1.5) 0

By the I.V.T such that ( ) 0 in the interval 1,1.5

g

g

g g

g x

(b) ( ) 2 cos( ) 1g x x x Using the formula

( ) ( )

( ) ( )

a g b b g ac

g b g a

Now using linear interplation on the interval

1<


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(1.0921615) 0.0060289 0

1.0921615<

cape pure mathematics unit 2module 2: sequences, series and approximations

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