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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 furtherpredict 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 confidencethat 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