Suppose you invest $\$2000$ in the bank. The money attracts an interest rate of $10\%$ per annum. The interest is added to the investment each year, so the total interest increases. The percentage increase each year is $10\%$, so at the end of the year, you will have $110\%$ of the value at its start. This corresponds to a multiplier of $1.1$.

After one year, it is worth: $\$2000 \times 1.1 = \$2200$

After two years it is worth: $\$2000 \times 1.1^2 = \$2420$

After three years it is worth: $\$2000 \times 1.1^3 = \$2662$

This suggests that if the investment is left for $n$ years it would be $\$2000 \times 1.1^{n}$.

Observe that:

$$ \begin{align} \displaystyle

u_{0} &= \$2000 &\text{initial investment}\\

u_{1} &= \$2000 \times 1.1 &\text{amount after 1 year}\\

u_{2} &= \$2000 \times 1.1^2 &\text{amount after 2 years}\\

u_{3} &= \$2000 \times 1.1^3 &\text{amount after 3 years}\\

u_{4} &= \$2000 \times 1.1^4 &\text{amount after 4 years}\\

\vdots \\

u_{n} &= \$2000 \times 1.1^n &\text{amount after } n \text{ years}\

\end{align} $$

## Example 1

$\$3000$ is invested for $6$ years at $44\%$ per annum compound interest, compound annually. What will it amount to at the end of this period? Give your answer to the nearest cent.

\( u_6 = \$3000 \times 1.04^6 = \$3795.9570 \cdots \)

The investment will amount to $\$3795.96$.

## Example 2

How much should I invest now if I need a maturing value of $\$20\ 000$ in $5$ years’ time, and I am able to invest at $7\%$ per annum compounded half-yearly ($6$ months)? Give your answer to the nearest cent.

The initial investment of $u_0$ is unknown.

The investment is compounded twice annually, so $r = 1 + \dfrac{0.07}{2} = 1.035$.

There are $5 \times 2 = 10$ compounding periods, so $n=10$.

\( \begin{align} \displaystyle

u_{10} &= u_{0} \times 1.035^{10} \\

\$20\ 000 &= u_{0} \times 1.035^{10} \\

u_{0} &= \dfrac{\$20\ 000}{1.035^{10}} \\

&= \$14\ 178.376 \cdots \\

&= \$14\ 178.38 \cdots \\

\end{align} \)

So I should invest $\$14\ 178.38$.

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