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

ℤ+

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

vf p f p

p p p

, stating the value of v.

(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 of S.

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

1

2 2 1 ( 1)2n

r n

r

r n

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

1(3 4) 3 11 .

2

n

r

r n n

4. (a) The expressions 26, 2 , and x x x form the first three terms of a

geometric progression. By calculating two different expressions for

the common ratio, form and solve an equation in x 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) and t u t u

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

(i) Use an algebraic method to show that one possible value of u is

5 and to find the other possible value of u.

(ii) For each possible value of u, calculate the value of the common

ratio of the series.

Given that 5u and that the sum to infinity of the geometric series is

896, calculate:

(iii) The value of t.

(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

(ii) The sum of the first 60 terms of the series.

6. (a) Find the first four terms of the of the sequence:

1 1

4, 7n n

u u u

(b) A sequence of terms {n

U }, 1n is defined by the recurrence relation

2 1

where is a constantn n n

U U U

Given also that1 2

2 and 5U U :

(i) Find an expression in terms of for 3

U

(ii) Find an expression in terms of for 4

U

Given that the value of4

21U :

(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 sequence 1 2 3 4, , , ,...u u u u is defined by

1 15 3(2 ), 7n

n nu u u

(a) Determine the first four terms of the sequence.

(b) Prove by mathematical induction fornℤ+, that 5 2n n

nu .

8. (a) Use Maclaurin’s theorem to find the first three non-zero terms in the

series expansion of (1 2 )

ln1 3

x

x

, and state the interval in x for

which the expansion is valid.

(b) (i) Show using Maclaurin’s theorem that

2

2 33( 3)sin3 3 3 ...

2

xe x x x x

where is a constant.

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

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

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

9. (a) Show that the Taylor expansion of sin( )x in ascending powers of

6

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

2

sin9

.

10. Given that 3cos( ) sin( ) 2 0

dyx y x y

dx and that 1y at 0x , use

Taylor’s method to show that, close to 0x , terms in 4x and higher powers

can be ignored,2 311 56

1 22 3

y x x x .

11. (a) Expand fully the expression 3(1 3 )(1 2 ) .x x

(b) Expand3(2 )y . Hence or otherwise, write down the expansion

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

(c) The coefficient of 2x in the expansion of

3(2 )(3 )x bx is 45. Find

the possible values of the constant b.

(d) Find the term independent of x in the expansion of

3

2 1.

2x

x

12. (a) Use the binomial series to expand 10

2 3x in ascending powers of x

up to and including the term in3x , giving each coefficient as an

integer.

(b) Use your series expansion, with suitable value for x, to obtain an

estimate for 1.9710, giving your answer to 2 decimal places.

13. (a) Find the binomial expansion of 2

1

x

x

in ascending powers of x as

far as the term in3x . State the range of values of x for which the

expansion is valid.

(b) Find an expansion of (1 2 )x up to and including the term in3x . By

substituting 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 of as far as the term in3 .

(iii) State the set of values of for which the expansion is valid.

14. (a) Evaluate9!

2!3!4!.

(b) Prove that

2! 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 n

r n rC C

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

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

intermediate value theorem.

(ii) Using the end points of this interval by interval bisection,

obtain a first and second approximation to x.

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

equation 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 to x. Give

your answer to three decimal places.

By Carlon R. Baird

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

( 1) ( 1)2

r r r r r

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

2 2r 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)

12

2

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

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

ℤ+

(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 1 1 ... ...

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

1 1

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)!

1 1

( 1)!

n

n

1 1( ) ( 1)

( 1) ( 1)( 1 1)

f p f pp p p p

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

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 lim 4( 1)(2 1)n n

n nS

n n

2

2

2

2

2

2 2

2

2 2 2

2

2 3lim

4 2 2 1

2 3lim

8 12 4

2 3

lim 8 12 4

32

lim 12 4

8

2 0

8 0 0

1

4

n

n

n

n

n n

n n n

n n

n n

n n

n nn n

n n n

n

n n

3. (a) Let Pnbe the statement

1

2 2 1 ( 1)2n

r n

r

r n

Showing 1

P 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

Verifying 1

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

1

P is truek

∴By Principle of Mathematical Induction Pnholds true n ℤ+

(b) Let Pnbe the statement

1

13 4 3 11

2

n

r

r n n

Showing 1

P is true:

L.H.S: 1

1

(3 4) 3(1) 4 7r

r

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 1

Pk

is 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

1

P is truek

∴By Principle of Mathematical Induction Pnholds true n ℤ+

4. (a)

From equ’n : 2x

ra

From equ’n : 2

2 xr

a

Substituting r into equ’n

2 2

2 2

2

2 2

2

4

4

x x

a a

x x

a a

x ax

From equ’n 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

(b) $a D

After 1 year: 4

$100

ar 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 $10,000

5 5

5

(1.04) $10,000

10000$ $8219.27

(1.04)

ar D

D

So Dylan’s initial investment was about $8219.27

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

1010

5

10000 1.04

(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 u

ar 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:

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

u

t 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 3

3(5) 1 16 4r

When 1

;2

u

12 2

1 222

3113 1

22

r

(iii) Given that 5u and 896S

(3(5) 1)896

311

4

16 896

1

4

16 224

14

a tS

r

t

t

t

(iv) 1

1

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)n

u a n d

205 (20 1)(4)

5 (19)(4)

81

u

(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 equ’ns:

2 47 a d

29 62 a d

Equ’n 62 12a d

Substituting a into equ’n

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

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 that 4

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 = 2

5

(c) 4 3

4 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 15

4 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

4 3

4 2

4 3

4 4 4

4 2

4 4

4

2

10 1lim lim

10 1

lim

1 110

lim 1

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 2n n

nu

Showing 1

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

1 1

1

1 1

1

5 2 7

P is true

u

u u

Assume k

P is true:

5 2k k

ku

Verifying k+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

1

P is truek

By Principle of Mathematical Induction P holds true n

n ℤ+

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

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 31 ln 4 ...

1 3 2 3

h x x x x

xx x x

x

where 1 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

(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 27

f e

2 3

By Maclaurin's theorem:

''(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

2

2 33 3

sin 3 3 3 ...2

xe x x x x

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

1'( ) (1 )

1q x x

x

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

q

q

q

2 3

2 3

2 3

2 2 3 3

By Maclaurin's theorem:

''(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 1

ln(1+ )= ...2 3

x x x x

Hence,

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

expansion of3sin 3 ln(1 ) is ,xe x x x x , this means that the co-

efficient of both x and x2 are 0.

2 0

2

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

2

2

2

1 3 0

2

13

2

1 43 ( 2) 2

2 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 | 1

sin6 6 2

f

'( ) cosf x x | 3

' cos6 6 2

f

''( ) sinf x x | 1

'' sin6 6 2

f

Using Taylor’s 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

(b)

10. (a) 3cos( ) sin( ) 2 0

dyx y x y

dx

Differentiating equ’n:

3

2

2

2

2

2

cos( ) sin( ) 2 0

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

cos( ) sin( )

d dy d dx y x y

dx 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 equ’n

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 0

1 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 18

1 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

3

0

0

0

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

0 2 0

2

dy

dx

dy

dx

dy

dx

Substituting values of 0

x ,0

0

and dy

ydx

into equ’n

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

dx

d y

dx

Substituting values of 0

x ,

2

0 2

0 0

, and dy d y

ydx dx

into equ’n

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 Taylor’s 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

Now substituting values:

2 3

2 3

2 3

0 01 2 0 11 112 ...

2! 3!

11 561 2 ...

2 3

Ignoring other coefficients:

11 561 2

2 3

x xy x

y x x x

y x x x

11. (a) Let’s first consider 3

1 2x

3 3 2 3 2 3

1 2

2 3

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

1 6 12 8

x C x C x x

x x x

3 2 3

2 3 2 3 4

4 3 2

1 3 1 2 1 3 1 6 12 8

1 6 12 8 3 18 36 24

24 44 30 9 1

x x x x x x

x x x x x x x

x x x x

(b) 3 3 0 2 1 1 2 0 33 3 3 3

0 1 2 32 2 2 2 2y C y C y C y C y

2 3

3 2

8 12 6

6 12 8

y y y

y y y

Hence

3 3 2

2 2 2 2

3 3 2 0 3 2 2 1 3 1 2 2 3 0 2 2

0 1 2 3

2 3 4 2

3 2 2 4 6 4 2

3 4 5 6 2 3 4 2

6

2 6 12 8

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

+ 6 2 12 12 8

3( ) ( ) 3( )( ) 6 12 12 8

3 3 6 12 6 12 12 8

3

x x x x x x x x

C x x C x x C x x C x x

x x x x x

x x x x x x x x x

x x x x x x x x x

x

5 3 211 6 12 8x x x x

(c) First of all, let’s expand 3

3 bx

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 0

3 (2 5) 3(2 5) 0

2 5 3 3 0

5 or 1

2

b b

b b

b b

b b b

b b b

b b

b b

(d)

3 The 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 of xobtaining an estimate for 10

1.97

2 3 1.97

3 2 1.97

0.01

x

x

x

3 0 1 2 33 2 1 0

2 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 13 3

2 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

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 1

2 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

where 1

12

x

12 3

2

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

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 16

1 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

491 2(0.01) 0.98

50

49 149

50 50

1 7

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

2

2

2

6+7 5 1 2 1 2 1 1

Let 1

6 7 1 5 1 0 2 2 3 2 0

18 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 C

B

A B C

A

A

2

2

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

2

2 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

(ii) ss

(iii) All expansions valid for |x|<1

1 1 1

1 2 3

2 3

2 3

1

( ) 2 1 3 1 4 2

Using binomial expansion:

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

2! 3!

= 2 1 ...

2 2 2 2 ...

Valid for 1

1 1 13 1 3 1 1

2

P

2 3

2 3

2 3

1

1

1

1

1 1 1 1 2...

! 3!

3 1 ...

3 3 3 3 ...

Valid for 1

1

14 2 4 2 1

2

1 4 2 1

2

1 2 1

2

1

1 2 3

2 3

2 3

1 1 1 1 1 1 1 21 1 1 12 1 2 1 1 ...

2 2 2! 2 3! 2

1 1 1 2 1 ...

2 4 8

1 1 2 ....

2 4

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

nn 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

(c) R.T.P: n n

r n rC C

i.e.

! !

!( )! ( )! !

n n

r n r n r n n r

! !

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

! =

!( )!

n n

n r n n rn 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

(ii)

2 3(2.5) 0.125 0

2 2

2< <2.5

2 2.52.25 2.359375 0

2

2.25< <2.5

2.25 2.5(2.375) 1.2285... 0

2

2.375< <2.5

2.4375 0.580

a bf f f

f f

f f

f

32... 0

2.4375< <2.5

2.46875 0.2348... 0

2.46875< <2.5

2.484375 0.05676651... 0

2.484375 < <2.5

2.4921875 0.0336604... 0

2.484375< <2.4921875

2.488

f

f

f

f

28125 0.011666... 0

2.48828125< <2.4921875

2.490234375 0.010968231... 0

2.48828125< <2.490234375

2.49 2 d.p.

f

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< <1.5

1| 0.78778 | 1.5 0.08060461.0464106...

0.78778 0.0806046

(1.0464106) 0.0478365... 0

1.0464106< <1.5

1.04641Now,

c

g

c

06 0.78778 1.5 0.04783651.0723772...

0.78778 0.0478365

(1.0723772) 0.0252732... 0

1.0723772< <1.5

Using linear interpolation on the new interval gives:

1.0723772 0.78778 1.5 0.0252732

0.78

g

c

1.0856695...778 0.0252732

(1.0856695) 0.0125400... 0

1.0856695< <1.5

1.0856695 0.78778 1.5 0.0125400Now, 1.0921615...

0.78778 0.0125400

g

c

(1.0921615) 0.0060289 0

1.0921615< <1.5

1.09 2 d.p.

g

x

17. 3 2( ) 3 5 4f x x x x

First of all, let’s find the derivative of the given function f(x).

2

1

'( ) 3 6 5

Using Newton Raphson's process:

'( )

n

n n

n

f x x x

f xx x

f x

Given the first approximation 0 1.4x

12 1

2

3 2

2

2

2

( )

'( )

(1.4) 1.4

'(1.4)

(1.4) (1.4) 3(1.4) 5 1.4 4

'(1.4) 3(1.4) 6(1.4) 5

0.136

0.136 1.4 1.45483...

2.48

1.455 3 d.p.

The second approximation 1.

f xx x

f x

f

f

f

f

x

x

455