Quadratic Patterns
● Pattern Structure: Sequence with a constant second difference.
● General Formula:
Tn = an² + bn + c
● Example: 3, 8, 15, 24, 35, ...
● Steps to Find the Formula:
1. First Differences:
-8-3=5
- 15 - 8 = 7
- 24 - 15 = 9
- 35 - 24 = 11
2. Second Differences (constant):
-7-5=2
-9-7=2
- 11 - 9 = 2
3. Since the second difference is constant, it’s a quadratic pattern.
○ Find a: a = (Second Difference) / 2 =
○ = 1.
○ A = 2ND DIFFERENCE ÷2
, 4. Set up equations using terms:
○ For T1 = 3
→ 1(1)² + b(1) + c = 3
→ 1 + b + c = 3.
→ b + c =2 ( EQUATION 1)
○ For T2 = 8:
→ 1(2)² + b(2) + c = 8
→ 4 + 2b + c = 8.
→ 2b +c =4 ( EQUATION 2)
○ Solve for b and c :
→ ALWAYS SUBTRACT EQUATION 1 FROM EQUATION 2
→ 2b + c =4
- b + c =2
—————
b =2
→ Substitute b=2 into equation 1 or 2 to find c
2(2)+c = 4
c=0
→b=2 and c = 0
→ Confirm this by Substituting
Tn = an^2 + bn + c
✔️
3= 1(1)^2 + 2(1)+0
3=3
5. Therefore the General Formula: Tn = 1n² + 2n + 0
● Pattern Structure: Sequence with a constant second difference.
● General Formula:
Tn = an² + bn + c
● Example: 3, 8, 15, 24, 35, ...
● Steps to Find the Formula:
1. First Differences:
-8-3=5
- 15 - 8 = 7
- 24 - 15 = 9
- 35 - 24 = 11
2. Second Differences (constant):
-7-5=2
-9-7=2
- 11 - 9 = 2
3. Since the second difference is constant, it’s a quadratic pattern.
○ Find a: a = (Second Difference) / 2 =
○ = 1.
○ A = 2ND DIFFERENCE ÷2
, 4. Set up equations using terms:
○ For T1 = 3
→ 1(1)² + b(1) + c = 3
→ 1 + b + c = 3.
→ b + c =2 ( EQUATION 1)
○ For T2 = 8:
→ 1(2)² + b(2) + c = 8
→ 4 + 2b + c = 8.
→ 2b +c =4 ( EQUATION 2)
○ Solve for b and c :
→ ALWAYS SUBTRACT EQUATION 1 FROM EQUATION 2
→ 2b + c =4
- b + c =2
—————
b =2
→ Substitute b=2 into equation 1 or 2 to find c
2(2)+c = 4
c=0
→b=2 and c = 0
→ Confirm this by Substituting
Tn = an^2 + bn + c
✔️
3= 1(1)^2 + 2(1)+0
3=3
5. Therefore the General Formula: Tn = 1n² + 2n + 0
