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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.

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Worksheets

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

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Recall — your gut answer first

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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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The investment recurrence relation

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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.

$$A_{n+1} = (1+r)A_n + a$$

$A_n$ 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.

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

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.

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), $a = \$200$:

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.

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 $A_1$.

$A_{n+1} = (1 + 0.006)A_n + 150 = 1.006A_n + 150, \quad A_0 = 2{,}000$

Identify $r = 0.006$ and $a = 150$. State $A_0$.

PROBLEM 2 · ITERATE TO A₄

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

$A_1 = 1.006(2{,}000) + 150 = \$2{,}162$

First step from $A_0$.

PROBLEM 3 · CLOSED FORM VERIFICATION

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

$A_4 = 2{,}000(1.006)^4 + 150 \times \dfrac{(1.006)^4 - 1}{0.006}$

Substitute into $A_n = A_0(1+r)^n + a \cdot \dfrac{(1+r)^n - 1}{r}$.

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?

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 ≠ closed form (they always match)

Students sometimes think a small discrepancy means they made an error. It does not — it is just rounding intermediate recurrence steps. Both methods are mathematically equivalent. State this in your answer if the values are close.

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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Work mode · how are you completing this lesson?

Quick-fire practice · 5 iterations

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

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

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

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

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

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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Multiple choice

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

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].

Boss battle · The Banker

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

Enter the arena

Science Jump · platform challenge

Climb platforms by answering recurrence relation questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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Module overview · Maths Advanced · Checkpoint 3