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Worked Solutions

1. A force acting on a particle at the point P, whose position vector, with respect to
an origin O, is . The torque of this force about O is calculated as: .
Given 3 5 2 and 7 3 3 and | | 2, determine the co‐ordinates of
the point P, assuming the force is measured in Newtons, is measured in metres and in
Nm.

Solution:

Let

| |
7 3
3 5
3 2
2 5 2 3 5 3
3 5 2

Requirements to be satisfied:

2 5 7… ∴ 1.5 1.5
2 3 3…....
5 3 3….. ∴ 3.5 2.5

and 4 | | 2

On substituting 1.5 1.5, 3.5 2.5 into 4

1.5 1.5 3.5 2.5 4, expand and solve quadratic 38 88 42 0

a =0.-0.495, b = -1.825, c= 0.67 OR

a = -0.9645, b=-0.6075, c=1.643

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2. Given:

z1  12(cos120  i sin120)
z 2  15(cos 20  i sin 20)
z 3  20(cos 230  i sin 230)
z 4  r(cos   i sin )

Find and  if z 3  z 4 2  z1  z 2 where 90    180

Solution:

20cis230 (r 2cis2)  12cis120 15cis(20)

 180cis140
r 2  9cis(90)  9cis270  r  3 and   135

3. If  is the nth root of unity, prove that :

n


k 1
k 1
0

Present a clear and valid mathematical argument.

Solution:

n  1 i.e.  is the nth root of unity.

n


k 1
k 1
 0  1  2  3  .......n 1

 1  w  2  3  .......n 1
1( n  1)
  0 since n  1
 1

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sin  (3 - tan 2 ) 
4. Remembering that tan  = , prove that tan 3 = tan   
cos  (1 - 3tan  ) 
2

Solution 2 methods

cos 3 + isin 3 = (cos  + isin )3 = cos3  + 3 cos2  isin  + 3 cos  i2sin2  + i3sin3 

cos 3 = cos3  ‐ 3 cos  sin2 , sin 3 = 3 cos2  sin  ‐ sin3 

Now tan3 = sin 3 / cos3

= (3 cos2  sin  ‐ sin3 ) / (cos3  ‐ 3 cos  sin2 )

= sin  (3 cos2  ‐ sin2 ) / cos  (cos2  ‐ 3 sin2 )

= tan  (3 ‐ sin2  / cos2 ) / (1‐ 3 sin2  / cos2 )

= tan  (3 ‐ tan2 ) / (1‐ 3 tan2 )

OR

LHE = (z3 – z -3)/2i (z3 + z -3)/2

 (z - z -1 ) 2   (z + z -1 )2 (z - z -1 )2 
 3 2  (z - z -1 )  3  
(z - z ) -1
i (z + z -1 )2   (z + z -1 )2 (z + z -1 )2  (z - z -1 )  3(z + z -1 )2  (z - z -1 )2 
RHE    
i(z + z -1 )  3(z - z -1 )2  i(z + z -1 )  (z + z )
-1 2
3(z - z -1 )2  i(z + z -1 )  (z + z -1 )2  3(z - z -1 )2 
1 
 i 2 (z + z -1 )2  
 (z + z -1 )2 (z + z -1 )2 
   
(z - z )  3(z + 2 + z )  (z - 2 + z )  (z - z )  4z + 4 + 4z 
-1 2 -2 2 -2 -1 2 -2
    
i(z + z -1 )  (z 2 + 2 + z -2 )  (z 2 - 2 + z -2 )  i(z + z -1 )  4z 2 - 4 + 4z -2 

(z - z -1 )  4z 2 + 4 + 4z -2  (z - z -1 )  z 2 + 1 + z -2   z 3 + z + z -1  z  z 1  z 3   z3  z 3 
      
i(z + z -1 )  4z 2 - 4 + 4z -2  i(z + z -1 )  z 2 - 1 + z -2   i(z3  z + z -1  z  z 1  z 3 )   i(z3  z 3 ) 
 LHE

5. When plotted on an Argand diagram and joined together the roots of a complex
7
equation form an equilateral triangle. One of the roots has an argument of  and
12
a modulus of 2. Write the original complex equation.

Solution:
The roots form an equilateral triangle so the original equation has power 3 and since

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7
z  2cis (  ) is one root
12
The original equation is

7 3
z 3  (2cis( )) = 4 2 (1  i )
12

3 2 
A 
6. Given 1 0  ,

 2n 1  1 2  2n 1 
Prove by mathematical induction that A n   n n 
, for n  1
 2 1 2  2 

Solution:

Step 1: Prove that the statement is true for n = 1.

 211  1 2  211  3 2

A  1
1
1 
  as required.
 2  1 2  2  1 0 
Step 2: Assume the statement is true for n = k

 2k 1  1 2  2k 1 
Ak   k k 
…………………………..(1)
 2  1 2  2 
Required to prove the statement is true for n = k +1

k 1  2k  2  1 2  2k  2 
A   k 1 k 1 
…………………………(2)
2 1 2  2 
Now

A k 1  A k  A
 2k 1  1 2  2k 1  3 2  3  2k 1  3  2  2k 1 2  2k 1  1(2) 
 k k    
 2  1 2  2  1 0   3 2  3  2  2 2  2k  2 
k k

 2  2k 1  1 2k  2  2 
 k 1 
 2  2  1 2  2 
k

 2k  2  1 2  2k  2 
  k 1 k 1 
as required
2 1 2  2 
The statement is therefore true for n = k+1

Step 3:

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Therefore, by the principle of mathematical induction the result holds for positive integer values
of k when n = k and when n = k+1. The result therefore holds for all positive integer values of n
provided n  1.

7.

8.

Select an origin, O, to form a right‐hand system of co‐ordinate axes, say, the left‐back corner of
the sandpit. Relative to this origin, the co‐ordinates of the vertices are respectively:, , and.
0,5,2 , 10,10,4 5,0,3 The shade cloth has an area represented by the ∆ .
This area may be represented by | |.

Now to find .
5 5
5 10 5 5 1
1 1 5 10 1
5 10 5 5 50 25
15 75

1 1 2
152 75 38.2 2
2 2

15 26
The area of the shade cloth is or approximately 38.2 m2.
2

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9.

(a) Assuming no air-resistance, the initial velocity r  0  is given by x iˆ  y ˆj m/s where


3
x  25cos  x  25   x  15
5
4
y  25sin   y  25   y  20
5

Hence, r  0   15iˆ  20 ˆj m/s



(b) At time t ,  r  t    g ˆj and selecting g  10 m / s 2 , leads to r  t     ˆj dt and hence


that r  t   10t ˆj  c .
 
Since r  0   15iˆ  20 ˆj then r  t   15iˆ   20  10t  ˆj m/s.
 

(c) (i) r  t   15iˆ   20  10t  ˆj implies that


r  t   15t iˆ   20t  5t 2  ˆj  c . The initial displacement of the
 
projectile may be expressed as r  0    ˆj and hence

the displacement at t time is given by
r  t   15t iˆ   5  20t  5t 2  ˆj . The time of flight of the

projectile can now be found by solving 5  20t  5t 2  0 .
The solutions being t  2  5 seconds. We reject t  2  5
because t  0 , hence t  2  5   4.236  seconds is the only
acceptable derived value of t. Hence the range is given by
 
15 2  5   63.541 metres from the foot of the launch platform.

 
(ii) Impact velocity is then determined from r 2  5  15iˆ  10 5 ˆj .

3 5
 
Hence r 2  5  5 29   26.926  m/s acting at tan 1   to
  10 
the downwards vertical ( i.e., at  3351 ).

(iii) Find the time when displacement is 25m.

r  t   15t iˆ   5  20t  5t 2  ˆj

| r (t ) ||15t iˆ   5  20t  5t 2  ˆj |

25  (15t ) 2   5  20t  5t 2 
2

252  (15t )2   5  20t  5t 2 

2

Using “solve” on GC gives 1

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v  t   15iˆ   20  10t  ˆj

Now v(1) = 15i + 10j
| v(1) |= 152  100  325  18.03m / s
Angle = tan 33.7 .

10.

(a) p1  t    a  2b  t iˆ + 2bt ˆj and p 2  t    2a  b  iˆ  b ˆj

 
p2  t     2a  b  iˆ  b ˆj.dt

p2  t    2a  b  t iˆ  bt ˆj + c
 
As p2  0   a ˆj , then p2  0   c and hence c  a ˆj
   
p2  t    2 a  b  t iˆ  bt ˆj + a ˆj

 p2  t    2a  b  t iˆ   a - bt  ˆj

(b) The particles collide when p1  t   p2  t 
 
That is, when  a  2b  t iˆ + 2bt ˆj   2 a  b  t iˆ   a - bt  ˆj
  a  2b  t   2a  b  t  A
and 2bt  a - bt  B

From equation A a  3b provided t  0 and from B 3bt  3b

provided that b  0 , produces t  1 second.
Therefore the particles collide after one second.

(c) The co-Ordinates of the point of intersection are found by evaluating either p1 1 or

p2 1 . In either case, the collision point is given by  a  2b  iˆ + 2b ˆj which reduces to

 5a 2a 
5b iˆ + 2b ˆj . Hence the point of intersection is  5b, 2b  provided b  0 or  , 
 3 3 
provided that a  0 .

11.

Locate the origin, O, at the foot of the altitude of the aircraft’s position at time t and in the
horizontal plane containing the stranded pilot. Assume that the canister is released at time t=0
seconds relative to O.

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The displacement of the canister at time t seconds after it is released is given by r  t  where the

unit vectors iˆ and ĵ are taken in the sense of + ĵ being vertically upwards and + iˆ horizontally
in the direction of motion relative to the origin, O.

From the given data, relative to the origin, O, r  0   120 ˆj and r  0   60 iˆ . We require the
 
displacement of the canister at time t i.e. at the point indicated by A in the above diagram.

Assuming there are no external forces acting on the motion of the canister, other than gravity,
then  r   g ˆj m / sec 2 and r  t    gt ˆj  c . Hence, r  t   60 iˆ  gt ˆj and from which
  
r  t   60t iˆ  21 gt 2 ˆj + c .
 

Applying the initial conditions, then we may say that r  t   60t iˆ    21 gt 2  ˆj represents

the location of the canister at time t. The canister lands in the water when r vert  t   0 , i.e.,
 
when 120  21 gt 2  0 .

Therefore, t  2 6 seconds ( if g is selected as 10 m / sec 2 ) and rejecting the negative value

of t since t  0 , then the range of the canister on impact with the water is given by 60t . From
which the range is determined as 120 6 metres.

The pilot is drifting at 1 m / sec because of the current. In 2 6 seconds, the pilot will have
drifted 2 6 metres.

Given that the range of the canister from the plane to the pilot is D metres, then
D  2 6  50  120 6 metres. Hence, D  118 6  50 metres and an acceptable range of
values for D is given by

118 6  50  D  118 6  50 metres.

However, the value D  118 6 metres should be excluded, as the canister will land on the
pilot !

12.

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13.

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14.

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15.

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16.

17.

18.

Write l1 in parametric form:

x  3  2k , y  2  3k , z  5  4k

Substitute the parametric equation into the equation for the plane.

2  3  2k   7  2  3k   3  5  4k   125
6  4k  14  21k  15  12k  125
23  37k  125
125  23
k
37
k 4
Determine the location on the line when k  4 .
x  3  2  4  11
y  2  3  4  10
z  5  4  4  11

The plane is intersected by the line at the point Q 11,10, 11 .

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The second particle travels through P  4, 6, 9 and Q 11,10, 11 .
d q p
  
  
d  11iˆ  10 ˆj  11kˆ  4iˆ  6 ˆj  9kˆ


d  15iˆ  16 ˆj  2kˆ

r akd
  
The path of the second particle is given by the vector equation,

  
r  4iˆ  6 ˆj  9kˆ  k 15iˆ  16 ˆj  2kˆ


or

  
r  11iˆ  10 ˆj  11kˆ  k 15iˆ  16 ˆj  2kˆ



19.
The system has infinitely many solutions. The solution can be described by the parametric equations of a line.

For this to be true;

nˆ1  nˆ2  nˆ3  0 and

nˆ1  nˆ2  nˆ2  nˆ3 and 0  nˆ2  nˆ3

iˆ ˆj kˆ
nˆ1  nˆ2  7 5 1
6 4 24
5 1 7 1
 iˆ  ˆj
4 24 6 24
7 5
 kˆ
6 4
 116iˆ  174 ˆj  58kˆ
The directional vector of the line of intersection of the three planes is given by,

116iˆ  174 ˆj  58 kˆ . As there are infinitely many solutions the free variable z will be defined as z  t .

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7 5 1 21 
 
A | B   6 4 24 156 
 0 0 0 0 
ref  A | B  :
1 0 2 12 
 
A | B  0 1 3 21 
0 0 0 0 

let z  t
From R 2 : y  3t  21
 y  21  3t
From R1 : x  2t  12
 x  12  2t
The line is defined by the parametric equations:

x  12  2t , y  21  3t , z  t
This solution must satisfy the third plane:

p3 : 2 x  b3 y  8 z  d3
Sub in x  12  2t , y  21  3t , z  t
2  12  2t   b3  21  3t   8t  d3
24  4t  21b3  3tb3  8t  d3
24  21b3  3tb3  12t  d3
For this to satisfy the equation
the terms containing the parameter
t must equal zero.
3tb3  12t  0
12t
b3   4
3t
Rewrite
24  21b3  3tb3  12t  d3
24  21  4  3t  4  12t  d3
60  d3
 d3  60
 p3 :  2 x  4 y  8 z  60

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20.

5  1
x  t    t   [1]
2 t 
 1
y  t   4  t   [2]
 t
Rewrite [1]:
2x 1
t
5 t
Squaring this gives,
4 x2 1
 t 2  2  2 [3]
25 t
Rewrite [2]:
y 1
t
4 t
Squaring this gives,
y2 1
 t 2  2  2 [4]
16 t
[3]  [4] :
4 x2 y 2
 4
25 16
x2 y 2
 1
25 64

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21.

a)

a  9.8

v   a dt
 
  9.8 dt
v  9.8t  c1
 
When t  0, v  0 :

 c1  0

 v  9.8t

x   v dt
 
  9.8t dt
x  4.9t 2  c2
 
When t  0, x  0 :

 c2  0

 x  4.9t 2

or

x  lt 2  mt  n
v  2lt  m
a  2l
It is known,
t  0 , v  0 m  0
a  9.8

2l  9.8
l  4.9
v  9.8t  m
v  9.8t
x  4.9t 2
Assume the well is at ground level,  n  0
t  6 , x  4.9  6
2

 176.4

The bottom of the well is 176.4 metres below ground level.

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b)

t  6,
v  9.8  6
v  58.8 m/s
Speed  v  58.8
 58.8 m/s

22.

a)
  360 revolutions per minute (rpm)
360  2
=
60
 12 rad s 1

b)
v  r
 12  12
 144 m sec 1  v  452.389 m sec 
1

23.

a)

v  23m/s   19
r  t   9.8 ˆj


r  t    9.8 ˆj dt

 9.8t  c1

r  0   23cos 19  iˆ  23sin 19  ˆj

 
r  t   23cos 19  iˆ  23sin 19   9.8t ˆj

r  0   1.7 ˆj

r  t   23t cos 19  iˆ

 
 1.7  23t sin 19   4.9t 2 ˆj

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Let the vertical component equal zero:
0  1.7  23t sin 19   4.9t 2
Solve for t using graphics calculator:
t  1.7289s (ignoring the negative solution)
r 1.7289   23 1.7289  cos 19  iˆ


 1.7  23 1.7289  sin 19   4.9 1.7289  ˆj
2

r 1.7289   37.5983iˆ

The ball travelled a horizontal distance of 40.6211 metres. The time of flight was 1.8679 seconds.

b) The ball will rise and reach maximum height when the vertical component of its velocity is zero.

 
r  t   23cos 19  iˆ  23sin 19   9.8t ˆj

23sin 19   9.8t  0
23sin 19 
t
9.8
t  0.7641s

r  0.7641  23  0.7641 cos 19  iˆ



 1.7  23  0.7641 sin 19   4.9  0.7641 ˆj
2

r  0.7641  16.6166iˆ  4.5608 ˆj


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24.

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i)
4rev  4s
1 rev  1s
2
T

2
1

  2 rads 1
ii )
Velocity of particle
v  r
v  0.75  2
v  1.5 ms 1
When water escapes the umbrella:
v H  1.5 ms 1 , vV  0 ms 1
Determine when water hits the
ground level.
x  lt 2  mt  n
v  2lt  m
a  2l
Gravity, a  9.8m/s 2
 l  4.9, m  0, n  1.8
x  4.9t 2  1.8
Let x  0
Solve for x:
1.8
x
4.9
x  0.6061s
The droplet travels 0.6061s at1.5 ms 1.
Distance =0.6061  1.5
 2.856 m
The radius of the circle:
r  
0.752  2.856 2 m
r  2.953 m
The diameter of the circle:
D  (2  r ) m
D  (2  2.953) m
D  5.906 m

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25.
5
     
 2cis  6   25 cis  5  
    6

   
3
 
4cis  43cis  3   
    3
  3 
32  5 
 cis    
64  6 
1  11 
 cis  
2  6 
1  
 cis   
2  6

26.

 z : z  2  3i  9
Substitute z  x  yi
x  yi  2  3i  9
 x  2    y  3 i 9

 x  2    y  3  9
2 2

 x  2    y  3  81
2 2

The equation represents a circle with a centre at C = (2, –3) and radius, r = 9.

27.

z4  1

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1  cis(0)

z 4  cis(0)
 0 2 k 
z  4 1cis   
4 4 

 2 0 
Let k  0, z1  cis  
 4 
 2k 
z  1cis    cis  0 
 4 
1

 2 1  
Let k  1, z2  cis  
 4 
 
 cis  
2

 2 2 
Let k  2, z3  cis  
 4 
 cis  

 2  3 
Let k  3, z4  cis  
 4 
 3 
 cis  
 2 

   3 
z1  1, z2 = cis   , z3 = cis   , z4 = cis  
2  2 

28.

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 Brisbane School of Distance Education
Solve 0  z  z  31z  111,
3 2

given z1  3 is a solution for P  z  .

z 2  2 z  37
z  3 z 3  z 2  31z  111
 ( z 3  3z 2   )
2 z 2  31z  111
( 2z2  6z  )
37 z  111
( 37 z  111)
0

0  z 2  2 z  37
0  z  2 z  1  1  37
0   z  1  36
2

0   z  1   36i 2 
2

0   z  1   6i 
2 2

0   z  1  6i  z  1  6i 
Using the Null factor law
z2  1  6i, z3  1  6i
The three roots to P  z  are given by
z1  3, z2  1  6i, z3  1  6i

29.

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a  11
za  z 3 
z
z  az  11  a  0
4 2

Let P  z   z 4  az 2  11  a
Determine a, given P  z   0


If  3i is a root of P  z  , then P  3i  0 
     a   3i 
4 2
P  3i   3i  11  a

P   3i   9  3a  11  a
9  3a  11  a  0
 4a  20  0
a5

P  z   z 4  5 z 2  11  5
 z 4  5z 2  6
P  a   a 2  5a  6 , where a  z 2
  a  3 a  2 
Let P  a   0
0   a  3 a  2 
a1  3 , a2  2
z 2  3 , z 2  2
z  3 , z  2
 z   3i,  2i,

30.

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Rearrange
d
z 2  5 z  9 
z
d
z 2  5z  9 
z
z  5z  9z  d  0
3 2

Substitute the first solution

z  2i
2  i  52  i  92  i  d  0
3 2

1  2   i   3  2   i   3  2   i   1 2   i 
3 0 2 1 1 2 0 3

 5  4  4i  i 2   18  9i  d  0
8  12i  6  i  20  20i  5  18  9i  d  0
5d  0
d 5
z 3  5z 2  9 z  5  0
z1  2  i, z2  z1  2  i
P  z    z  a  z  2  i  z  2  i  , a  R
  z  a   z 2  4 z  5
z 3  5 z 2  9 z  5   z  a   z 2  4 z  5
Hence,
 5  5a
5
a
5
a 1
P  z    z  1 z  2  i  z  2  i 
The solutions to P  z   0 are,
z1  2  i, z2  2  i, z3  1

31.

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3 1
n 1,  1 . Therefore true for n  1
2
Assume true for n  k

3k  1
1  3  9    3k 1 
2
Consider n  k  1

3k  1 k
1  3  9    3k 1  3k  3
2
3k  1 2  3k
 
2 2
3 1  2   1
k


2
k 1
3 1

2

Therefore true for n  k  1 if true for n  k . True for n  1 therefore true  n  N .

32.
If n  1

LHS = 2

RHS  2  1  2 Therefore true for n  1 .

Assume true for n  k

k

  4r  2   2k
r 1
2

Consider n  k  1
k 1 k

  4r  2     4r  2   4  k  1  2
r 1 r 1

 2 k 2  4k  2
 2  k 2  2k  1

 2  k  1
2

Therefore true for n  k  1 if true for n  k . True for n  1 therefore true n  1 .

33.

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Consider n  1

 2 n  1
2
 1  8 Therefore true for n  1

Assume true for n  k

 2k  1
2
 1  8m
4k 2  4k  8m
Consider n  k  1

 2  k  1  1  1  4  k  1  4  k  1  1  1
2 2

 4  k 2  2k  1  4k  4
 4k 2  8k  4  4k  4
 4k 2  4k  8k  8
 8m  8k  8
 8  m  k  1
Therefore true for n  k  1 if true for n  k . True for n  1 , therefore true for all n  N ..

34.

Consider n  1

13  4  9 Therefore true for n  1

Assume true for n  k

13k  4k  9m
Consider n  k  1

13k 1  4k 1  13 13k  4  4k
  9  4  13k  4  4k
 9 13k  4  13k  4  4k
 9 13k  4 13k  4k 
 9 13k  4  9m 
 9 13k  4m 
Therefore true for n  k  1 if true for n  k . True for n  1 , therefore true for all n  1 .

35.

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Consider n  0 :

3n2  32  9

103  1  101  1  9
n

Therefore true for n  0

Assume true for n  k

103  1  3k  2 m
k

103  3k  2 m  1
k

Consider n  k  1
k 1
103  1  103 3  1
k

  1
3
 103
k

  3k  2 m  1  1
3

  3k  2  m3  3  3k  2  m 2  3  3k  2  m  1  1
3 2

 33k  6 m3  32 k 5 m 2  3k 3 m
 3k 332 k 3 m3  3k 33k  2 m 2  3k 3 m
 3k 3  32 k 3 m3  3k  2 m 2  m 
k 1
k3 k 2
Therefore 103  1 is divisible by 3 if 103  1 is divisible by k
k
. The statement is true for
n  0 , therefore true  n  0 .

IA2 general revision ‐ 2020