hi guys!
i was wodnering if you could please help me tackle this question from the 2014 hsc paper. i dont know why i just found it confusing and i legit dont know what to do
thank you very much
Hey! You want to start by building up the series. Let the balance at the end of the month be \(A_n\). We step through what happens to the account balance in each month:
- Add $500 at the start
- Apply interest at the end
So at the end of the first month we have:
)
Now the next month, we don't add $500. We add 1% more, \(500(1.01)\). So now we have this:
\right)(1.003)\\A_2=500(1.003)^2+500(1.01)(1.003))
This is Part (i) done!Next month, we don't add \(500(1.01)\). We add 1% greater AGAIN, so, \(500(1.01)^2\). So now we have:
^2\right)(1.003)\\A_3=500(1.003)^3+500(1.01)(1.003)^2+500(1.01)^2)(1.003))
By now we should notice the pattern. Jump to \(A_n\):
^n+500(1.003)^{n-1}(1.01)+...+500(1.003)^2(1.01)^{n-2}+500(1.003)(1.01)^{n-1}\\A_n=500\left((1.003)^n+(1.003)^{n-1}(1.01)+...+(1.003)^2(1.01)^{n-2}+(1.003)(1.01)^{n-1}\right))
Now what is inside the brackets is actually a
very messy geometric series. We could define it in various ways by pulling out different terms, but I'll define it as a series with first term \(a=1.003^n\), and common ratio:
We get this by looking at what happens to the powers of (1.003) and (1.01) - The powers of the former go down, so we are dividing by it. The powers of the latter go up, so we are multiplying by it, hence this value of \(r\)!
Besides this, this is just a standard question. Use the formula for the sum of a geometric series:
}{\frac{1.01}{1.003}-1})
Now thankfully, we don't need to rearrange this monstrosity - Just set \(n=60\) and calculator:
To the nearest cent
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