Name: Mishka
Surname: Habib Ismail
Student Number: 66941113
Unique Number: 560736
Assignment: 3
Module: OPM1501
1
,PART A
1.) 0 -4
1 1
X5 -4
2 6
k
5k - 4
2.)
Input 0 1 2 k
values
Output -4 1 6 5k – 4
values
3.) 0, 3, 8, 15, 24
The next term in this sequence is 24.
3.1.)
0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120
+7 +9 +11 +13 +15 +17 +19 +21
+3 +5
There is a second difference of 2.
The eleventh term of the sequence is 120.
3.2.) Terms are formed and these terms formed are natural numbers.
2
, 4.)
1; 2; 3; 4; … and tn= n + 1
1; 9; 17; 25; … and tn= 8n - 7
1; 8; 15; 22; 29; … and tn= 7n - 6
1; 7; 13; 19; … and tn= 6n – 5
5.) Yes, Lester is correct.
This has been checked and his claim is accurate.
The sum of the two numbers (the one which is on the left and the one which is on the
right) is twice the number in the middle.
The sum of the two numbers (the one which is at the top and the one which is on the
left) is also twice the number in the middle.
Therefore, the sum of all outside numbers (top, bottom, left and right) are four times
the middle number.
E.g. 6 + 12 + 20 + 14 = 52 and
the middle number is 13.
13 x 4 = 52
Also
14 + 12 = 26
6 + 20 = 26
26 + 26 = 52
3
Surname: Habib Ismail
Student Number: 66941113
Unique Number: 560736
Assignment: 3
Module: OPM1501
1
,PART A
1.) 0 -4
1 1
X5 -4
2 6
k
5k - 4
2.)
Input 0 1 2 k
values
Output -4 1 6 5k – 4
values
3.) 0, 3, 8, 15, 24
The next term in this sequence is 24.
3.1.)
0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120
+7 +9 +11 +13 +15 +17 +19 +21
+3 +5
There is a second difference of 2.
The eleventh term of the sequence is 120.
3.2.) Terms are formed and these terms formed are natural numbers.
2
, 4.)
1; 2; 3; 4; … and tn= n + 1
1; 9; 17; 25; … and tn= 8n - 7
1; 8; 15; 22; 29; … and tn= 7n - 6
1; 7; 13; 19; … and tn= 6n – 5
5.) Yes, Lester is correct.
This has been checked and his claim is accurate.
The sum of the two numbers (the one which is on the left and the one which is on the
right) is twice the number in the middle.
The sum of the two numbers (the one which is at the top and the one which is on the
left) is also twice the number in the middle.
Therefore, the sum of all outside numbers (top, bottom, left and right) are four times
the middle number.
E.g. 6 + 12 + 20 + 14 = 52 and
the middle number is 13.
13 x 4 = 52
Also
14 + 12 = 26
6 + 20 = 26
26 + 26 = 52
3