## 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.

#### 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.

#### 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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