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 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, 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 \( \$14\ 178.38 \).
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