Use second differences and derive a finite geometric sum
Recover a quadratic nth term from constant second differences · Derive and apply the finite geometric-series formula · Handle r = 1 separately
Section Paper 2 · Historical Extended · non-calculator
Exam scope and assessment notes
Extended content; not a Core requirement. Historical study of Cambridge IGCSE Mathematics (US) 0444, 2025 Version 2. The final assessment was June 2025; the syllabus was available only to schools in Arizona, USA. This is not a current exam-entry option. Original self-check teaching and explained practice are not Cambridge papers, official marks or predicted grades. Linked teaching foundations do not establish full syllabus mastery. Calibrated cumulative assessment, reliable drawn-work review and qualified mathematics-teacher review remain unfinished.
Before you begin
- Arithmetic and geometric nth terms
- Solving simultaneous linear equations
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What should you establish before applying the method in this lesson?
What you will learn
- Recover a quadratic nth term from constant second differences
- Derive and apply the finite geometric-series formula
- Handle r = 1 separately
Quadratic pattern, stated as a model
For uₙ = An² + Bn + C at consecutive integer n, the second difference is 2A. Once A is known, substitute two terms to find B and C. Constant second differences support the assumed quadratic model; finite data alone do not prove an infinite rule.
Subtract a shifted sum
Let Sₙ = a + ar + … + arⁿ⁻¹. Multiplying gives rSₙ = ar + … + arⁿ. Subtracting cancels the middle terms: (1 − r)Sₙ = a(1 − rⁿ). For r ≠ 1, Sₙ = a(1 − rⁿ)/(1 − r). The equivalent a(rⁿ − 1)/(r − 1) gives the same result.
When r = 1, every term is a and Sₙ = na. Never divide by 1 − r at r = 1. A finite sum needs no condition |r| < 1; that condition belongs to an infinite-sum limit, which is not needed here.
Sources for this lesson
Pause and explain
What is the sum of seven terms all equal to 4?
Worked example
Find the total of six payments 5, 10, 20, …, assuming each doubles.
Show the worked solution
- a = 5, r = 2, n = 6.
- S₆ = 5(2⁶ − 1)/(2 − 1) = 5×63 = 315.
- Direct terms 5 + 10 + 20 + 40 + 80 + 160 confirm 315.
Answer 315 payment units.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Assuming a quadratic pattern, find uₙ for 2, 6, 12, 20, … and predict u₅.
Give me a hint
Compute first and second differences.
Compare my reasoning
- Differences are 4, 6, 8; second differences are 2, so A = 1.
- 1 + B + C = 2 and 4 + 2B + C = 6 give B = 1, C = 0.
- uₙ = n² + n; u₅ = 30.
uₙ = n² + n; u₅ = 30 under the quadratic assumption.
Look for these in your work
- Use consecutive indexing
- Check all supplied terms
Find the sum of the first five terms of 12, 6, 3, …, assuming a geometric sequence.
Give me a hint
Use a = 12 and r = 1/2; there are five terms.
Compare my reasoning
- S₅ = 12(1 − (1/2)⁵)/(1 − 1/2).
- 12×(31/32)×2 = 93/4.
93/4 = 23.25.
Look for these in your work
- Count the terms correctly
- Keep exact fractions until the end
A grant pays 100 in year 1 and grows by 10% each year for three years. Find the total; compare with no annual increase.
Give me a hint
A 10% increase has ratio 1.1, not 0.1.
Compare my reasoning
- Payments are 100, 110, 121; their sum is 331.
- The formula gives 100(1.1³ − 1)/(1.1 − 1) = 331.
- With r = 1, the total is 3×100 = 300.
331 with growth; 300 with no increase.
Look for these in your work
- Model the multiplier
- Separate total from final payment
- Use the r = 1 case correctly
Common mistakes
- A sum of n terms is not the nth term.
- The geometric-sum formula for r ≠ 1 cannot be substituted at r = 1.
What can you explain now?
What is the sum of seven terms all equal to 4?
Compare with the explanation
28.
The ratio is 1, so use na = 7×4.
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Make it stick
What is the sum of seven terms all equal to 4?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.