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) Open Quadratic equations in new window. 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

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


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


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 \(P(1-r)^n\).


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


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 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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