Compound Interest/Examples/Arbitrary Example 1
Jump to navigation
Jump to search
Example of Compound Interest
Let $\pounds 1000$ be invested for $2$ years at $8 \%$ per annum.
Let interest be compounded half-yearly.
At the end of the $1$st $6$ months, the compound interest will be:
- $I_1 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1000 = \pounds 40$
At the end of the $2$nd $6$ months, the compound interest will be:
- $I_2 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1040 = \pounds 41 \cdotp 60$
At the end of the $3$rd $6$ months, the compound interest will be:
- $I_3 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1081 \cdotp 60 = \pounds 43 \cdotp 26$
At the end of the $4$th $6$ months, the compound interest will be:
- $I_4 = \dfrac 8 {100} \times \dfrac 1 2 \times \pounds 1124 \cdotp 86 = \pounds 44 \cdotp 99$
Hence the total interest earned will be $\pounds 169 \cdotp 85$.
Proof
Using Formula for Compound Interest:
- $I = P \paren {\paren {1 + r}^n - 1}$
In this case:
- $P = 1000$
- $n = 4$
- $r = \dfrac {8 \%} 2 = 4 \% = 0.04$
Hence:
- $I = 1000 \paren {\paren {1 + 0 \cdotp 04}^4 - 1} \approx 1000 \times \paren {1 \cdotp 16985 - 1} = 169 \cdotp 85$
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): interest
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): interest