Compound Interest | Series and Sequences
Compound Interest is being used to calculate the total investment over time. Suppose John invests $1000 in the bank. He leaves the money in the bank for four years and is paid an interest rate of 10% per annum. The interest is added to his investment yearly, so the total value increases.
The percentage increase each year is 10%, so at the end of the year, John will have \( 100\% + 10\% = 110\% \) of the value at its start. This corresponds to a multiplier of 1.1.
After one year, the investment is worth;
\( $1000 \times 1.1 = $1100 \)
After two years, the investment is worth;
\( $1000 \times 1.1^{2} = $1210 \)
After three years, the investment is worth;
\( $1000 \times 1.1^{3} = $1331 \)
After four years, the investment is worth;
\( $1000 \times 1.1^{4} = $1464.10 \)
Practice Questions
Question 1
\( $1000 \) is invested for five years at 12% per annum compound interest, compounded annually. What will the amount to at the end of this period?
\( \begin{aligned} \displaystyle
u_5 &= $5000 \times 1.12^{5} \\
&= $8811.71
\end{aligned} \)
Question 2
\( $1000 \) is invested for five years at 12% per annum compound interest, compounded monthly. What will the amount to at the end of this period?
\( \begin{aligned} \displaystyle
u_5 &= $5000 \times 1.01^{5 \times 12} \\
&= $9083.48
\end{aligned} \)
Question 3
How much should I invest now if the maturing value of \( $10 \ 000 \) is in 5 years? The investment is at 12% per annum compounded quarterly.
\( \begin{aligned} \displaystyle
u_1 \times 1.03^{5 \times 4} &= $10 \ 000 \\
u_1 &= \dfrac{$10 \ 000}{1.03^{5 \times 4}} \\
&= $5536.76
\end{aligned} \)
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