Betty received an SMS from a friend regarding a Bitcoin (BTC) investment that guarantees a fixed percentage of interest per month. Betty clicked on the link in the SMS and found out that the BTC investment has an account for each investor. An interest is added to the account at the end of each month at a fixed rate of 11% of the amount in the account at the beginning of the month.

(i)
Betty decides to start an investment account with $1500 on 1 Jan 2022. She invests with the intention of not withdrawing any money out of the account but just leave it for the interest to build up. Find the total profit at the end of one full year.
(ii)
Betty forwards the SMS to her friend, Carl who decides to invest $200 on the first day of each month starting from 1 Jan 2022.

Show that the projected amount of money in Carl's investment account at the end of $n$ months of investment is given by
$\frac{22200}{11}\left[\left(\frac{111}{100}\right)^n - 1\right].$

To determine the amount of special coating to apply to the exterior of the spherical part of the light bulb, the manufacturer wishes to find the surface area of the spherical part of the light bulb.

It is given that the area of a surface of revolution formed by revolving the graph of $x = \mathrm{f}(y)$ around the $y$ -axis over the interval from $y = a$ to $y = b$ is
$\int_{a}^{b} 2\pi \mathrm{f}(y) \sqrt{1 + \left(\mathrm{f}'(y)\right)^2} \mathrm{d}y.$

Find numerical answer of the surface area of the spherical part of the light bulb.
(iii)
Find the projected amount of money in Carl's investment account at the end of one year.

Each light bulb can effectively light up a circular area of $0.5\text{ m}^2$ . The art student wants to estimate the number of light bulbs needed for the art installation in part (a). The art student uses the following formula to calculate the number of light bulbs to acquire.
$\text{Number of light bulbs} = \frac{\text{Area of Region } R}{0.5}$

Given that 1 unit on the art installation represents $100\text{ m}$ , determine the number of light bulbs that the art student would acquire. Suggest why the number of light bulbs using this formula may not be accurate.

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Worked solution

11(i)

1. Calculate the total amount in the account at the end of 12 months.

Since an interest of $11\%$ is added to the previous month's balance, the amount at the end of each month forms a geometric progression with a common ratio of $1.11$ .
$\text{Amount after 12 months} = 1500 \times \left(1 + \frac{11}{100}\right)^{12}$
$= 1500(1.11)^{12}$
$\approx 5247.6779$

2. Calculate the total profit by subtracting the initial principal.

$\text{Total profit} = \text{Final Amount} - \text{Initial Investment}$
$= 5247.6779 - 1500$
$= 3747.6779$
$\approx \$3747.68 \text{ (to 2 d.p.)}$

Answer

$3747.68

11(ii)

1. Express the value of each deposit at the end of n months

A monthly interest of 11% multiplies the amount by $\dfrac{111}{100}=1.11$ each month.

The $200 deposited on 1 Jan earns interest for $n$ months: $200(1.11)^n$ .

The $200 deposited on the first day of month 2 earns interest for $n-1$ months: $200(1.11)^{n-1}$ .

$\;\vdots$

The $200 deposited on the first day of month $n$ earns interest for $1$ month: $200(1.11)^{1}$ .

Total $=200(1.11)+200(1.11)^2+\cdots+200(1.11)^n$ .

2. Recognise the GP and apply the sum formula

This is a geometric series with first term $a=200(1.11)$ , common ratio $r=1.11=\dfrac{111}{100}$ and $n$ terms.

$S_n=\frac{a(r^n-1)}{r-1}=\frac{200\left(\frac{111}{100}\right)\left[\left(\frac{111}{100}\right)^n-1\right]}{\frac{111}{100}-1}.$

3. Simplify to the required closed form

$\frac{111}{100}-1=\frac{11}{100}.$

$S_n=200\cdot\frac{111}{100}\cdot\frac{100}{11}\left[\left(\frac{111}{100}\right)^n-1\right]=\frac{200\times111}{11}\left[\left(\frac{111}{100}\right)^n-1\right]$

$=\frac{22200}{11}\left[\left(\frac{111}{100}\right)^n-1\right].\quad\text{(shown)}$

Answer

Total amount $=\dfrac{22200}{11}\left[\left(\dfrac{111}{100}\right)^n-1\right]$ (shown).

11(iii)

1. Identify the value of $n$ for one year

One full year of monthly investments corresponds to $n = 12$ .

2. Substitute into the formula from part (ii) and evaluate

Using the result from part (ii):
$\frac{22200}{11}\left[\left(\frac{111}{100}\right)^{12} - 1\right]$
$= \frac{22200}{11}\big[(1.11)^{12} - 1\big]$
$= 2018.18\ldots \times (3.49845\ldots - 1)$
$= 2018.18\ldots \times 2.49845\ldots$
$\approx \$5042.31$

Answer

Approximately $5042.31 (i.e. about $5042.30).

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Worked solution

11(i)

11(ii)

11(iii)

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