# Compound Interest of Investment

Suppose you invest $\2000$ in the bank. The money attracts an interest rate of $10 \%$ per annum. The interest is added to the investment yearly, so the total interest increases. Each year’s percentage increase 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 for6\$ 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, and I can 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 $\14\ 178.38$.