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hscscience Maths Adv · Y12
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Module 7 · L11 of 20 ~40 min ⚡ +95 XP available

Recurrence Relations for Investments

Every time you check your bank balance you are reading the output of a recurrence relation, a rule that takes yesterday's balance, adds interest, adds your deposit, and produces today's balance. In this lesson you'll learn to write, read, and solve these relations: the mathematical engine powering every savings account, super fund, and investment portfolio on Earth.

Today's hook, Your bank processes millions of accounts every night using exactly the formula $A_{n+1} = (1+r)A_n + a$. That single line of mathematics is also why starting a savings plan at 20 versus 30 can be worth hundreds of thousands of dollars by retirement.
0/5QUESTS

Get oriented

Recall what you already know, meet the key ideas and settle the terms.

Worksheets

Practise this lesson

Three printable worksheets that build from foundations to mastery, or build your own from any module’s questions.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Build custom Build your own from any module question
01
Recall, your gut answer first
+5 XP warm-up

A savings account starts with $\$1,000$. Each month it earns 0.5% interest, and you deposit an extra $\$100$.

Month 0: $1,000  |  Month 1: $1,005 + $100 = $1,105  |  Month 2: $1,105 × 1.005 + $100 = $1,210.53

Without calculating further predict whether Month 3 will be approximately $1,315. Explain your reasoning.

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02
The investment recurrence relation
+5 XP to read

A recurrence relation describes how a quantity changes from one step to the next. For an investment with regular contributions there are only two things happening each period: the balance earns interest, then a new deposit arrives.

The master formula for any investment with regular deposits is:

Read it as: Next balance = current balance with interest + new deposit.

Each period the balance grows by the rate, then the regular deposit is added.
$$A_{n+1} = (1+r)A_n + a$$
$A_n$ and $r$
$A_n$ = balance at period $n$. $r$ = interest rate per period (as a decimal, 0.5% = 0.005).
$a$ = contribution
$a$ is the regular payment added each period. If no deposits are made, set $a = 0$.
Starting condition
Always state $A_0$, the initial balance. Every subsequent $A_n$ depends on it.
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What you'll master
Know

Key facts

  • The recurrence relation $A_{n+1} = (1+r)A_n + a$
  • How to build a step-by-step balance table
  • The link between recurrence and annuity formulas
Understand

Concepts

  • Why recurrence relations model real accounts perfectly
  • The difference between recursive and closed-form solutions
  • How compounding and contributions interact period by period
Can do

Skills

  • Write recurrence relations from word problems
  • Calculate balances iteratively step by step
  • Verify closed-form answers with recurrence steps
  • Compare investment strategies using recurrence tables
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Key terms
Recurrence relationA rule expressing each term in a sequence using previous terms: $A_{n+1} = (1+r)A_n + a$.
Initial conditionThe starting value $A_0$ needed to anchor a recurrence relation and generate the sequence.
Interest rate per period$r$ expressed as a decimal; e.g. 0.4% per month = 0.004.
Regular contribution$a$, the fixed deposit made at the end of each period.
Closed-form solution$A_n = A_0(1+r)^n + a \cdot \dfrac{(1+r)^n - 1}{r}$, jumps directly to any period $n$.
IterationApplying the recurrence rule one step at a time, building the sequence period by period.

Iterate the recurrence

Step an investment balance forward one period at a time.

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Iterating the recurrence relation
core concept

The power of the recurrence relation is that you don't need any new formula, you just plug in the previous answer and crank the handle. Let's trace $A_0 = \$5{,}000$, $r = 0.004$ (0.4% per month), $a = \$200$:

Step-by-step iteration table of an investment recurrence relation across periods 0 to 3.

Each row feeds the next. Period 3 uses Period 2's balance as its input.

$$A_{n+1} = (1+r)A_n + a \qquad \text{where } A_0 \text{ is the starting balance}$$
Why banks use recurrence. Every night, banking software runs this loop across millions of accounts, apply interest, add direct debits and credits, produce tomorrow's balance. The recurrence relation is not just a textbook construct; it is the live, running code of the financial system.

Investment recurrence: $A_{n+1} = (1+r)A_n + a$ where $r$ = rate per period (decimal), $a$ = deposit per period; Always state $A_0$ (initial balance), the whole sequence depends on it

Pause, copy the investment recurrence $A_{n+1} = (1+r)A_n + a$ where $r$ = rate per period and $a$ = deposit per period, always stating $A_0$ as the starting condition, into your book.

Did you get this? True or false: in the recurrence relation $A_{n+1} = (1+r)A_n + a$, the variable $a$ represents the interest rate per period.

Work the examples

Follow three worked recurrence problems end to end.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · WRITE THE RECURRENCE

An investment account starts with $\$2{,}000$. It earns 0.6% per month and $\$150$ is deposited at the end of each month. Write the recurrence relation and find $A_1$.

1
$A_{n+1} = (1 + 0.006)A_n + 150 = 1.006A_n + 150, \quad A_0 = 2{,}000$
Identify $r = 0.006$ and $a = 150$. State $A_0$.
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PROBLEM 2 · ITERATE TO A₄

Using $A_0 = \$2{,}000$, $r = 0.006$, $a = \$150$, find $A_4$ by iteration.

1
$A_1 = 1.006(2{,}000) + 150 = \$2{,}162$
First step from $A_0$.
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PROBLEM 3 · CLOSED FORM VERIFICATION

Verify $A_4 = \$2{,}653.85$ using the closed-form formula.

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$A_4 = 2{,}000(1.006)^4 + 150 \times \dfrac{(1.006)^4 - 1}{0.006}$
Substitute into $A_n = A_0(1+r)^n + a \cdot \dfrac{(1+r)^n - 1}{r}$.
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Quick check: In the closed-form formula $A_n = A_0(1+r)^n + a \cdot \dfrac{(1+r)^n - 1}{r}$, what does the second term represent?

Avoid the traps

Fix the off-by-one and starting-balance slips that cost marks.

Common errors · the 3 traps that cost marks

Trap 01
Wrong rate period
An annual rate of 6% is not 0.06 per month. Always match the rate to the contribution period. Monthly → divide annual rate by 12. Using the annual rate with monthly contributions will massively overstate the balance.
Trap 02
Recurrence vs closed form: they match exactly
The closed form is the recurrence unrolled, so the two are the same calculation and agree to the cent. If your two answers differ, you rounded an intermediate value and carried it into the next line, so go back and keep more decimals. Do not write off a gap as rounding unless you can say how much rounding would account for and check that it matches what you see, because a difference bigger than that is an arithmetic slip.
Trap 03
Forgetting $A_0$
A recurrence relation without an initial condition is incomplete. You must state both the rule and $A_0$. HSC markers expect both, one mark can hang on a missing starting condition.

Think through this: An account earns 4.8% p.a. and receives monthly deposits. What monthly interest rate $r$ should you use in the recurrence relation?

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Drill it and revisit

Run the drill, then return to your opening answer and name what has changed.

Quick-fire practice · 5 iterations

1

$A_0 = \$1{,}000$, $r = 0.005$, $a = \$100$. Find $A_1$.

2

$A_0 = \$3{,}000$, $r = 0.004$, $a = \$0$. Find $A_2$.

3

Write the recurrence relation: $A_0 = \$5{,}000$, monthly rate 0.3%, monthly deposit $\$250$.

4

An annual rate of 6% with monthly deposits: what is $r$ per month?

5

$A_0 = \$2{,}000$, $r = 0.006$, $a = \$150$. Find $A_3$.

Fill in the blanks: The closed-form solution for a recurrence relation investment is $A_n = A_0(1+r)^n$ + $a$ × [blank]. The first term represents the growth of the [blank] and the second term is the future value of all [blank].

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Match each scenario to the correct description:

  • $a = 0$
  • $r = 0$
  • $A_0 = 0$
  • Both $a = 0$ and $A_0 = 0$
  • Balance stays zero forever
  • No initial balance, contributions build from zero
  • No interest, balance grows only by deposits
  • No regular contributions, pure compound growth
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Revisit your thinking

Earlier you predicted whether Month 3 would be approximately $\$1,315$. The answer: $A_3 = 1.005 \times 1{,}210.53 + 100 = \$1{,}316.58$. Your prediction was very close. The pattern looks almost linear in early months because interest earned is still small relative to the $100 deposit. As time passes, interest compounds on a larger base and the growth curve bends upward, accelerating beyond what any linear prediction can capture.

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Practise investment recurrences

Write full short-answer responses, then check them against the model answers.

01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 43 marks

Q1. An account has $A_0 = \$5{,}000$, earns 0.4% per month, and receives $\$200$ per month. (a) Write the recurrence relation. (b) Find $A_1$ and $A_2$ by iteration. (3 marks)

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ApplyBand 43 marks

Q2. $A_0 = \$3{,}000$, monthly rate 0.6%, monthly deposit $\$150$. (a) Find $A_3$ using the closed-form formula. (b) Briefly explain the difference between the recurrence and closed-form approaches. (3 marks)

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AnalyseBand 54 marks

Q3. $A_0 = \$5{,}000$, $r = 0.005$ per month, $a = \$500$ per month. (a) Write the recurrence relation and find $A_1$, $A_2$, $A_3$. (b) Explain why the balance grows faster than linearly as $n$ increases. (4 marks)

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Comprehensive answers (click to reveal)

Drill 1: $A_1 = 1.005(1{,}000)+100 = \$1{,}105$

Drill 2: $A_1 = 1.004(3{,}000)+0 = \$3{,}012$; $A_2 = 1.004(3{,}012)+0 = \$3{,}024.05$

Drill 3: $A_{n+1} = 1.003A_n + 250$, $A_0 = 5{,}000$

Drill 4: $r = 6\%/12 = 0.5\% = 0.005$ per month

Drill 5: $A_1 = 1.006(2{,}000)+150 = \$2{,}162$; $A_2 = 1.006(2{,}162)+150 = \$2{,}324.97$; $A_3 = 1.006(2{,}324.97)+150 = \$2{,}488.92$

Q1 (3 marks): (a) $A_{n+1} = 1.004A_n + 200$, $A_0 = 5{,}000$ [1]. (b) $A_1 = 1.004(5{,}000)+200 = \$5{,}220$ [1]; $A_2 = 1.004(5{,}220)+200 = \$5{,}440.88$ [1].

Q2 (3 marks): (a) $A_3 = 3{,}000(1.006)^3 + 150 \times [(1.006)^3-1]/0.006 = 3{,}054.32 + 452.71 = \$3{,}507.03$ [2]. (b) Recurrence builds step-by-step using the previous balance; closed form jumps directly to any period $n$ using an explicit formula [1].

Q3 (4 marks): (a) $A_{n+1} = 1.005A_n + 500$, $A_0 = 5{,}000$ [1]; $A_1 = \$5{,}525$, $A_2 = \$6{,}052.63$, $A_3 = \$6{,}582.89$ [2]. (b) The balance grows faster than linearly because interest compounds on both the growing principal and all previous contributions, the interest earned each period increases, adding an accelerating component on top of the regular deposit [1].