Sequences and Series

[Morrice]

Solution seems wrong, so I'm curious to what the answer may be.
I got $42 509.53

[RuiAce]

Yeah, their solution has a mistake. However your's has a minor error as well.

You wrote that \( S_{18} = 1000\left( \frac{1.02^{4\times 19}-1}{1.02-1} \right) \), but it should in fact be \( S_{18} = 1000 \left( \frac{1.02^{4\times 19}-1}{1.02^4-1} \right) \) - the common ratio is \(1.02^4\) so the denominator also has that ^4 in it.

Regardless, thanks for bringing this up. You're most certainly correct in that there should be nineteen terms in the sum, instead of 18. I will forward a message to my manager.

Edit: You're also right in that it should be 1000, not 500.

Edit #2: Upon further glance, it looks like your final answer was correct. I think you knew what you were doing and just didn't write it down correctly but no biggie - message is passed on.

[Morrice]

Yeah, their solution has a mistake. However your's has a minor error as well.

You wrote that \( S_{18} = 1000\left( \frac{1.02^{4\times 19}-1}{1.02-1} \right) \), but it should in fact be \( S_{18} = 1000 \left( \frac{1.02^{4\times 19}-1}{1.02^4-1} \right) \) - the common ratio is \(1.02^4\) so the denominator also has that ^4 in it.

Regardless, thanks for bringing this up. You're most certainly correct in that there should be nineteen terms in the sum, instead of 18. I will forward a message to my manager.

Edit: You're also right in that it should be 1000, not 500.

Edit #2: Upon further glance, it looks like your final answer was correct. I think you knew what you were doing and just didn't write it down correctly but no biggie - message is passed on.

Thank you for reviewing. Yes I approached the problem from a different angle and was lazily writing parts of it next to the solution for contrast, only to miss out on the ^4.

While we're at it, there's also a slight error in the solution to the very first question of the same test, if you don't mind checking it out.
Thanks again.