Compound Interest

Suppose you invest $$2000$ in the bank. The money attract an interest rate of $10$% per annum. The interest is added to the investment each year, so the total interest increases.

Compound Interest Formula

Compound Interest Problems

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.

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.