Investing by regular instalments

The future value of a superannuation fund or annuity with regular deposits, using both the accountant's method (summing each deposit's growth) and the recursive method.

The accountant's method

Track each deposit separately, work out how long it earns interest, then add the results as a GP.

Example 1: superannuation●●●

$\$3000$ is deposited at the start of each year for 25 years, at $6\%$ p.a. compound interest. Find the final value.

The first deposit earns interest for 25 years and grows to $3000 \times 1.06^{25}$. The last earns for 1 year: $3000 \times 1.06$.

Each deposit

$A = 3000(1.06 + 1.06^2 + \dots + 1.06^{25})$

Add them

$A = 3000 \times \dfrac{1.06(1.06^{25} - 1)}{0.06} \approx \$174\,469.15$

Sum the GP

The recursive method

Let $A_n$ be the balance after $n$ years. Each year, add the deposit and then the interest: $$A_1 = 3000 \times 1.06,\quad A_2 = (A_1 + 3000) \times 1.06 = 3000(1.06^2 + 1.06),\ \dots$$ The pattern gives the same GP as the accountant's method.

Example 2: monthly deposits●●

$\$200$ is deposited at the end of each month for 10 years at $6\%$ p.a. compounded monthly. Find the final value.

The monthly rate is $0.5\%$, over $120$ months. The last deposit earns nothing, the first earns $119$ months.

$A = 200(1 + 1.005 + \dots + 1.005^{119}) = 200 \times \dfrac{1.005^{120} - 1}{0.005} \approx \$32\,775.87$

Flashcards

Deposits at the start of each period

$M(r + r^2 + \dots + r^n)$

Deposits at the end of each period

$M(1 + r + \dots + r^{n-1})$

Accountant's method

Follow each deposit separately, then sum the GP.

Practice

$\$3000$ at the start of each year for 25 years at $6\%$ p.a. Find the final value to the nearest cent.
$\$200$ at the end of each month for 10 years at $6\%$ p.a. compounded monthly. Find the final value to the nearest cent.
In Example 1, the first deposit grows to:
For deposits at the end of each year, the last deposit grows to:
$\$1000$ is deposited at the end of each year at $10\%$ p.a. Find the balance just after the second deposit.
$\$1000$ is deposited at the start of each year at $10\%$ p.a. Find the balance at the end of the second year.
The recursive method builds the balance by:
Write $1 + 1.05 + 1.05^2 + \dots + 1.05^{n-1}$ as $\dfrac{1.05^n - 1}{d}$. What is $d$, as a percentage?
Which earns more: deposits at the start of each year, or at the end (same amounts, rate and years)?