Superannuation Question (1 Viewer)

Stabilo123 · Jan 26, 2012

Jenny puts aside $20 at the end of each month for 3 years. How much will she have at the end of the 3 years if the investment earns 8.2% per annum paid monthly.

Could someone show me full working please

Carrotsticks · Jan 26, 2012

$\\ $Consider the first month. She puts in 20 dollars at the end of the month, but she won't get any interest just yet (she only JUST deposited!)$ \\\\ $We can work out quite simply that her monthly compound interest will be:$ \\\\ \frac{8.2}{12} \div 100 = \frac{41}{6000} $ percent per month.$$

$\\$Denote the amount of money she has after $ n $ months by $ T_n \\\\ \therefore T_1 = 20 \\\\ $After 1 month, she will have had some interest added, but she will also have deposited 20 dollars. This means that:$ \\\\ \begin{align*} T_2 &= 20 \times \left (1+ \frac{41}{6000} \right ) + 20 \\ &= 20 \left (\frac{6041}{6000} \right ) + 20 \end{align*} \\\\ $Now, we will apply interest to the second month (to find how much she has in the third month!), but also remember to add 20 dollars:$ \\\\ \begin{align*} T_3 &= T_2 \times \left (\frac{6041}{6000} \right ) + 20 \\ &=\left (20 \left (\frac{6041}{6000} \right ) + 20 \right ) \times \left (\frac{6041}{6000} \right ) + 20 \ \\ &= 20 \left (\frac{6041}{6000} \right )^{2} + 20\left (\frac{6041}{6000} \right ) + 20 \\ &= 20 \left ( \left (\frac{6041}{6000} \right )^{2} + \left (\frac{6041}{6000} \right ) + 1 \right ) \end{align*} \\\\$

$\\ $From here, we see a very obvious pattern occurring. If it goes up to the third month, then the power of $ \left (\frac{6041}{6000} \right ) $ is 2, so we can deduce that if it goes up to the $ n^{th} $ month, then the power goes up to $ n-1. \\\\ $Following this trend, we make the series: $ \\\\ T_n = 20 \left ( \left (\frac{6041}{6000} \right )^{n-1} + \left (\frac{6041}{6000} \right )^{n-2} + ... + 1 \right ) \\\\\\ $There are 36 months in a year, so it goes up to $ T_{36}: \\\\ T_{36} = 20 \left ( \left (\frac{6041}{6000} \right )^{35} + \left (\frac{6041}{6000} \right )^{34} + ... + 1 \right ) \right ) \\\\\\ $We see that the expression inside the big bracket is a Geometric Series, so we can use the 'Sum of G.P Formula':$ \\\\$

$\begin{align*} T_{36} &= 20 \left ( \frac{ \left ( \frac{6041}{6000} \right )^{36}-1}{ \left ( \frac{6041}{6000} \right )-1} \right ) \\\\ &\approx 813.16 \end{align*} \\\\ $Just as a note of interest (pun intended), if she were to invest 20 dollars per month for 36 months (no interest), she would have $ 36 \times 20 = 720 $ dollars, meaning that she earns approximately $ 813-720 = 93 $ dollars interest.$$

Carrotsticks · Jan 27, 2012

I just realised that I said "There are 36 months in a year" lol.

36 months in 3 years obviously.

jimmysmith560 · Aug 22, 2019

It's 2019 and I searched up the exact same question. Can't believe I actually found the working out. Thanks a lot.

Superannuation Question (1 Viewer)

Stabilo123

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Carrotsticks

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Carrotsticks

Retired

jimmysmith560

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