Recurrence Relation Calculator
VCE General Maths Units 1 & 2 (Sequences and finance): enter V0, R and D to see every term worked out, step by step.
Work out R from an interest or depreciation rate
| n | Working | Vn | Change |
|---|
Values are shown to 2 decimal places, but each step uses the full unrounded value, just like your CAS.
Reading a recurrence relation
A recurrence relation is a rule that tells you how to get the next value from the current one. It always has two parts: a starting value, and a rule.
V0 = 2500, Vn+1 = Vn + 100
V0 is where you start (the value at time zero). Vn is the value after n steps, so V3 is the value after three years, months or thousand kilometres. The rule above says "to find the next value, add 100 to the current value". When a sequence isn't about money, you'll often see tn used instead of Vn. It works exactly the same way.
The two types in Units 1 & 2
Linear (arithmetic)
Add or subtract the same amount, D, each step.
Vn+1 = Vn + D
Rule for term n: Vn = V0 + n × D
Used for simple interest (growth), and flat rate and unit cost depreciation (decay). Graph: points in a straight line.
Geometric
Multiply by the same amount, R, each step.
Vn+1 = R × Vn
Rule for term n: Vn = Rn × V0
Used for compound interest (R = 1 + r/100) and reducing-balance depreciation (R = 1 − r/100). Graph: points on a curve.
Where this fits in your course
| Section | Topic | Try this example |
|---|---|---|
| 3B | Writing recurrence relations in symbolic form | Any example: read the relation in the blue box |
| 3C–3D | Arithmetic sequences and recursion | Set R = 1 and choose your own D |
| 3E | Simple interest, flat rate and unit cost depreciation | Simple interest, Flat rate, Unit cost |
| 3F–3G | Geometric sequences and recursion | Set D = 0 and choose your own R |
| 3H | Compound interest and reducing-balance depreciation | Compound interest, Reducing balance |
| 3I | Finding term n with a rule | Check the "rule for term n" line against the table |
Section references are to Chapter 3, Sequences and finance, in Cambridge Senior Mathematics General Mathematics VCE Units 1 & 2 (2nd edition).
Worked example
Priya invests $4500 at 5.5% per annum, compounding yearly.
a) Write a recurrence relation for the value of the investment. b) Find the value after 1, 2 and 3 years. c) Use a rule to find the value after 10 years.
- Define the variable. Let Vn be the value of the investment after n years.
- Find R. R = 1 + 5.5/100 = 1.055
- Write the relation. V0 = 4500, Vn+1 = 1.055 × Vn
- Generate the terms.
V1 = 1.055 × 4500 = $4747.50
V2 = 1.055 × 4747.50 = $5008.61
V3 = 1.055 × 5008.6125 = $5284.09 - Use the rule for term n. V10 = 1.05510 × 4500 = $7686.65
- After 10 years the investment is worth $7686.65. Load the "Compound interest" example above to check every step.
Common mistakes
- Using the rate instead of R. 5.5% interest means R = 1.055, not 0.055 or 5.5. A 15% depreciation means R = 0.85, because 85% of the value is left.
- Forgetting to split the rate. 6% per annum compounding monthly is 6 ÷ 12 = 0.5% per month, so R = 1.005 and n counts months. The helper above does this for you.
- Working out D for simple interest from the wrong value. Simple interest is always calculated on the original principal, so D = r/100 × V0 every year.
- Getting the sign of D wrong. Depreciation takes value away, so D is negative in the rule (for example Vn+1 = Vn − 2400).
- Counting steps from 1. V0 is the start. After 3 years you want V3, not V4.
- Rounding as you go. Keep full values in your calculator and round only the final answer, or you can drift a few cents off.
Looking ahead to Unit 3
In Year 12 you'll combine both types into one rule, Vn+1 = R × Vn + D. This models loans, annuities and investments with regular payments, where interest is added and a payment is made each step. The three "Unit 3 preview" examples above let you try it now.