## Content

### Applications to finance

One of the many applications of sequences and series occurs in financial mathematics. Here we will briefly discuss compound interest and superannuation.

#### Compound interest

Compound interest was discussed in the TIMES module Consumer arithmetic (Year 9) . We invest an amount $\(P\) at an interest rate \(r\) paid at the end of a prescribed interval for a period of \(n\) such intervals. Interest rates are usually quoted as percentages, and so an interest rate of 3% means that \(r = 0.03\).

Let us here assume the time interval is one year. Hence, after one year, the investment has the value \(P+rP = P(1+r)\). After the end of two years, the investment has the value \(P(1+r) + rP(1+r) = P(1+r)^2\). We can see from this that, after \(n\) years, the investment will be worth \(P(1+r)^n\).

These amounts form a geometric sequence with common ratio \(1+r\), and the \(n\)th term is

\[ P(1+r)^n. \]#### Example

If $1000 is invested at 4% p.a. compounded yearly, what is the value of the investment after ten years?

#### Solution

Here \(P = 1000\), \(r = 0.04\) and \(n = 10\). Thus, after ten years, the investment is worth

\[ 1000(1 + 0.04)^{10} \approx \$1480.24. \]Depreciation is closely related to compound interest. When a company, for example, buys a car for work-related purposes, it is able to claim the depreciation in the value of the car over time as a tax deduction.

If the car is initially worth $\(P\) and is depreciated at a rate of \(r\) per annum, then the value of the car after one year is \(P-rP = P(1-r)\). After the end of two years, the car has value \(P(1-r) - rP(1-r) = P(1-r)^2\). We can see from this that, after \(n\) years, the car will be worth \(P(1-r)^n\).

#### Example

What is the value of a car after ten years, if it is initially worth $30 000 and is depreciated at 6% p.a.?

#### Solution

Here \(P = 30\,000\), \(r = 0.06\) and \(n = 10\). Thus, after ten years, the car is worth

\[ 30\,000\, (1 - 0.06)^{10} \approx \$16\,158. \]#### Superannuation

Superannuation is a way of saving for retirement. Money is regularly invested over a long period of time and (compound) interest is paid. Suppose I invest in a superannuation scheme for 30 years which pays 6% per annum. I put $3000 each year into the scheme, and (for the sake of simplicity) we will suppose that the interest is added yearly.

Thus, the first $3000 will be invested for the full 30 years at 6% per annum. Hence, it will accrue to \(\$3000 \times 1.06^{30}\). The next $3000, deposited at the beginning of the second year, will be invested for 29 years and so will accrue to \(\$3000 \times 1.06^{29}\), and so on. Writing the amounts from smallest to largest, the total value of my investment after 30 years is

\[ A = 3000 \times 1.06 + 3000 \times 1.06^2 + \dots + 3000 \times 1.06^{30}. \]This is a geometric series with first term \(a = 3000 \times 1.06\) and common ratio \(r = 1.06\); the number of terms is \(n = 30\). Using the formula for the sum of a finite geometric series, the final value of the investment is

\[ A \approx \$251\,405. \]This example illustrates both the value of regular saving and the power of compound interest. The $90 000 invested becomes roughly $251 400 over 30 years.

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