1. ## 2U math help

2014 Mathematics 2U HSC paper, question 16 b):
At the start of a month, Jo opens a bank account and makes a deposit of $500. At the start of each subsequent month, Jo makes a deposit which is 1% more than the previous deposit. At the end of every month, the bank pays interest of 0.3% (per month) on the balance of the account. (i) Explain why the balance of the account at the end of the second month is$500 (1.003)^2 + \$500 (1.01) (1.003) .
(ii) Find the balance of the account at the end of the 60th month, correct to
the nearest dollar.

I can do (i), but (ii) just looks so hard, because you have one number (1.01) increasing in power, but then (1.003) decreasing in power, it feels so complicated. Can someone explain this question as simple as possible?

2. ## Re: 2U math help

Well I hope you can see that

$A_{60} = 500[(1.003)^{60} + (1.003)^{59}(1.01) + \dots + (1.01)^{59}(1.003)]$

You can take a factor of $1.003^{60}$ to get

$500(1.003)^{60}[1+ \frac{1.01}{1.003} + \frac{1.01^2}{1.003^2} + \dots + \frac{1.01^{59}}{1.003^{59}}]$

From there it's a sum of a geometric series with a ratio of $\frac{1.01}{1.003}$

3. ## Re: 2U math help

omg I sort of get it now! how did you know to factorise that out to get that pattern???

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