# 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 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$n=10. \begin{align} \displaystyle u_{10} &= u_{0} \times 1.035^{10} \\ \20\ 000 &= u_{0} \times 1.035^{10} \\ u_{0} &= \dfrac{\20\ 000}{1.035^{10}} \\ &= \14\ 178.376 \cdots \\ &= \14\ 178.38 \cdots \\ \end{align} So I should invest\$14\ 178.38$. 