MIP1502 ASSESSMENT 3 2025
Question 1
1.1.1.1
Small triangles:
4 − 1; 9 − 4; 16 − 9; 25 − 16; 36 − 25
3; 5; 7; 9; 11
Black triangles:
3 − 1; 5 − 3; 7 − 5; 9 − 7; 11 − 9
2; 2; 2; 2; 2
Grey triangles:
1 − 0; 3 − 1; 6 − 3; 10 − 6
1; 2; 3; 4
White triangles:
0 − 0; 1 − 0; 3 − 1; 6 − 3; 10 − 6
0; 1; 2; 3; 4
1.1.1.2
Small triangles: the pattern is quadratic.
Black triangles: the pattern is linear.
Grey triangles: the pattern is quadratic.
White triangles: the pattern is quadratic.
,1.1.1.3
Small triangles have a constant second difference, so the pattern is quadratic.
Black triangles have a constant first difference, so the pattern is linear.
1.1.2.1
Grey triangles:
Starting from 0, add 1 then 2, then 3, then 4, then 5, and so on.
To obtain the number of grey triangles on the (𝑛 + 1)th diagram from the 𝑛th diagram, you
take the number of grey triangles on the 𝑛th diagram, then add 𝑛.
White triangles:
Starting from 0, add 0 then 1, then 2, then 3, then 4, and so on.
To obtain the number of white triangles on the (𝑛 + 1)th diagram from the 𝑛th diagram, you
take the number of white triangles on the 𝑛th diagram, then add (𝑛 − 1).
1.1.2.2
Grey triangles:
Recursive formula: 𝑇𝑛+1 = 𝑇𝑛 + 𝑛, 𝑇1 = 0
𝑇2 = 𝑇1 + 1 = 0 + 1 The last term is 1 and 1 = 2 − 1 (Sum has 2 terms)
𝑇3 = 𝑇2 + 2 = 0 + 1 + 2 The last term is 2 and 2 = 3 − 1 (Sum has 3 terms)
𝑇4 = 𝑇3 + 3 = 0 + 1 + 2 + 3 The last term is 3 and 3 = 4 − 1 (Sum has 4 terms)
𝑇5 = 𝑇4 + 4 = 0 + 1 + 2 + 3 + 4 The last term is 4 and 4 = 5 − 1 (Sum has 5 terms)
So, on 𝑇𝑛 the last term on the sum with 𝑛 terms is (𝑛 − 1).
𝑇𝑛 = 0 + 1 + 2 + ⋯ + (𝑛 − 3) + (𝑛 − 2) + (𝑛 − 1)
𝑇𝑛 = (𝑛 − 1) + (𝑛 − 2) + (𝑛 − 3) + ⋯ + 2 + 1 + 0
𝑇𝑛 + 𝑇𝑛 = [0 + (𝑛 − 1)] + [1 + (𝑛 − 2)] + [2 + (𝑛 − 3)] + ⋯ + [(𝑛 − 3) + 2]
+ [(𝑛 − 2) + 1] + [(𝑛 − 1) + 0]
2𝑇𝑛 = [𝑛 − 1] + [𝑛 − 1] + [𝑛 − 1] + ⋯ [𝑛 − 1] + [𝑛 − 1] + [𝑛 − 1]
2𝑇𝑛 = 𝑛 × [𝑛 − 1]
, 𝑛 𝑛2 𝑛
𝑇𝑛 = 2 [𝑛 − 1] OR 𝑇𝑛 = −2
2
𝑛 → [÷ 2] → [× (𝑛 − 1)]
White triangles:
Recursive formula: 𝑇𝑛+1 = 𝑇𝑛 + (𝑛 − 1), 𝑇1 = 0
𝑇2 = 𝑇1 + (1 − 1) = 0 + 1 − 1 = 0 1 term
𝑇3 = 𝑇2 + (2 − 1) = 0 + 1 − 1 + 2 − 1 = 0 + 1 + 2 − 2 = 0 + 1 2 terms
𝑇4 = 𝑇3 + (3 − 1) = 0 + 1 − 1 + 2 − 1 + 3 − 1 = 0 + 1 + 2 + 3 − 3 = 0 + 1 + 2 3 terms
𝑇5 = 𝑇4 + (4 − 1) = 0 + 1 − 1 + 2 − 1 + 3 − 1 + 4 − 1 = 0 + 1 + 2 + 3 + 4 − 4 = 0 + 1
+2 + 3 4 terms
So, on 𝑇𝑛 the last term on the sum with (𝑛 − 1) terms is (𝑛 − 2).
𝑇𝑛 = 0 + 1 + 2 + ⋯ + (𝑛 − 4) + (𝑛 − 3) + (𝑛 − 2)
𝑇𝑛 = (𝑛 − 2) + (𝑛 − 3) + (𝑛 − 4) + ⋯ + 2 + 1 + 0
𝑇𝑛 + 𝑇𝑛 = [0 + (𝑛 − 2)] + [1 + (𝑛 − 3)] + [2 + (𝑛 − 4)] + ⋯ + [(𝑛 − 4) + 2]
+ [(𝑛 − 3) + 1] + [(𝑛 − 2) + 0]
2𝑇𝑛 = [𝑛 − 2] + [𝑛 − 2] + [𝑛 − 2] + ⋯ [𝑛 − 2] + [𝑛 − 2] + [𝑛 − 2]
2𝑇𝑛 = [𝑛 − 1] × [𝑛 − 2]
[𝑛 − 1]
𝑇𝑛 = [𝑛 − 2]
2
𝑛 → [−1] → [÷ 2] → [× (𝑛 − 2)]
Question 1
1.1.1.1
Small triangles:
4 − 1; 9 − 4; 16 − 9; 25 − 16; 36 − 25
3; 5; 7; 9; 11
Black triangles:
3 − 1; 5 − 3; 7 − 5; 9 − 7; 11 − 9
2; 2; 2; 2; 2
Grey triangles:
1 − 0; 3 − 1; 6 − 3; 10 − 6
1; 2; 3; 4
White triangles:
0 − 0; 1 − 0; 3 − 1; 6 − 3; 10 − 6
0; 1; 2; 3; 4
1.1.1.2
Small triangles: the pattern is quadratic.
Black triangles: the pattern is linear.
Grey triangles: the pattern is quadratic.
White triangles: the pattern is quadratic.
,1.1.1.3
Small triangles have a constant second difference, so the pattern is quadratic.
Black triangles have a constant first difference, so the pattern is linear.
1.1.2.1
Grey triangles:
Starting from 0, add 1 then 2, then 3, then 4, then 5, and so on.
To obtain the number of grey triangles on the (𝑛 + 1)th diagram from the 𝑛th diagram, you
take the number of grey triangles on the 𝑛th diagram, then add 𝑛.
White triangles:
Starting from 0, add 0 then 1, then 2, then 3, then 4, and so on.
To obtain the number of white triangles on the (𝑛 + 1)th diagram from the 𝑛th diagram, you
take the number of white triangles on the 𝑛th diagram, then add (𝑛 − 1).
1.1.2.2
Grey triangles:
Recursive formula: 𝑇𝑛+1 = 𝑇𝑛 + 𝑛, 𝑇1 = 0
𝑇2 = 𝑇1 + 1 = 0 + 1 The last term is 1 and 1 = 2 − 1 (Sum has 2 terms)
𝑇3 = 𝑇2 + 2 = 0 + 1 + 2 The last term is 2 and 2 = 3 − 1 (Sum has 3 terms)
𝑇4 = 𝑇3 + 3 = 0 + 1 + 2 + 3 The last term is 3 and 3 = 4 − 1 (Sum has 4 terms)
𝑇5 = 𝑇4 + 4 = 0 + 1 + 2 + 3 + 4 The last term is 4 and 4 = 5 − 1 (Sum has 5 terms)
So, on 𝑇𝑛 the last term on the sum with 𝑛 terms is (𝑛 − 1).
𝑇𝑛 = 0 + 1 + 2 + ⋯ + (𝑛 − 3) + (𝑛 − 2) + (𝑛 − 1)
𝑇𝑛 = (𝑛 − 1) + (𝑛 − 2) + (𝑛 − 3) + ⋯ + 2 + 1 + 0
𝑇𝑛 + 𝑇𝑛 = [0 + (𝑛 − 1)] + [1 + (𝑛 − 2)] + [2 + (𝑛 − 3)] + ⋯ + [(𝑛 − 3) + 2]
+ [(𝑛 − 2) + 1] + [(𝑛 − 1) + 0]
2𝑇𝑛 = [𝑛 − 1] + [𝑛 − 1] + [𝑛 − 1] + ⋯ [𝑛 − 1] + [𝑛 − 1] + [𝑛 − 1]
2𝑇𝑛 = 𝑛 × [𝑛 − 1]
, 𝑛 𝑛2 𝑛
𝑇𝑛 = 2 [𝑛 − 1] OR 𝑇𝑛 = −2
2
𝑛 → [÷ 2] → [× (𝑛 − 1)]
White triangles:
Recursive formula: 𝑇𝑛+1 = 𝑇𝑛 + (𝑛 − 1), 𝑇1 = 0
𝑇2 = 𝑇1 + (1 − 1) = 0 + 1 − 1 = 0 1 term
𝑇3 = 𝑇2 + (2 − 1) = 0 + 1 − 1 + 2 − 1 = 0 + 1 + 2 − 2 = 0 + 1 2 terms
𝑇4 = 𝑇3 + (3 − 1) = 0 + 1 − 1 + 2 − 1 + 3 − 1 = 0 + 1 + 2 + 3 − 3 = 0 + 1 + 2 3 terms
𝑇5 = 𝑇4 + (4 − 1) = 0 + 1 − 1 + 2 − 1 + 3 − 1 + 4 − 1 = 0 + 1 + 2 + 3 + 4 − 4 = 0 + 1
+2 + 3 4 terms
So, on 𝑇𝑛 the last term on the sum with (𝑛 − 1) terms is (𝑛 − 2).
𝑇𝑛 = 0 + 1 + 2 + ⋯ + (𝑛 − 4) + (𝑛 − 3) + (𝑛 − 2)
𝑇𝑛 = (𝑛 − 2) + (𝑛 − 3) + (𝑛 − 4) + ⋯ + 2 + 1 + 0
𝑇𝑛 + 𝑇𝑛 = [0 + (𝑛 − 2)] + [1 + (𝑛 − 3)] + [2 + (𝑛 − 4)] + ⋯ + [(𝑛 − 4) + 2]
+ [(𝑛 − 3) + 1] + [(𝑛 − 2) + 0]
2𝑇𝑛 = [𝑛 − 2] + [𝑛 − 2] + [𝑛 − 2] + ⋯ [𝑛 − 2] + [𝑛 − 2] + [𝑛 − 2]
2𝑇𝑛 = [𝑛 − 1] × [𝑛 − 2]
[𝑛 − 1]
𝑇𝑛 = [𝑛 − 2]
2
𝑛 → [−1] → [÷ 2] → [× (𝑛 − 2)]
