Q8) As soon as you graduate you start thinking of retirement. At the end of your first year after school you invest 4,000 into your IRA account and every year thereafter you increase that contribution by 3%, starting in year 2. If you're able to keep this up for a total of 45 years, what will be your balance at the end of year 45? Assume you invest in an index fund that give you a 9% return per year.

Round your answer to the nearest dollar. For example, if your answer is $100.68, please enter $101.

(Please show your work in Excel and enter the numerical answer below)

Question

Q8) As soon as you graduate you start thinking of retirement. At the end of your first year after school you invest 4,000 into your IRA account and every year thereafter you increase that contribution by 3%, starting in year 2. If you're able to keep this up for a total of 45 years, what will be your balance at the end of year 45?  Assume you invest in an index fund that give you a 9% return per year.

Round your answer to the nearest dollar. For example, if your answer is $100.68, please enter $101.

(Please show your work in Excel and enter the numerical answer below)

BlackTom AI · Accepted Answer

We’re asked to find the balance after 45 years given a starting annual contribution of 4000 at the end of year 1, with contributions growing 3% each year thereafter, and an annual investment return of 9%.

First, identify the cash flow pattern:
- Year 1 contribution: C1 = 4000
- Year t contribution: Ct = 4000 * (1.03)^{t-1} for t = 1,...,45
- Each contribution is invested for (45 - t) years at 9% per year, so it accumulates by a factor of (1.09)^{45 - t} by end of year 45.

Thus, the balance at end of year 45 is:
Balance = sum_{t=1}^{45} [ Ct * (1.09)^{45 - t} ]
       = 4000 * sum_{t=1}^{45} [(1.03)^{t-1} * (1.09)^{45 - t}]

To simplify, factor out common terms to use a geometric-series form. Note that:
(1.03)^{t-1} * (1.09)^{45 - t} = (1.09)^{45} * (1.03)^{t-1} * (1.09)^{-t} = (1.09)^{45} * (1.03/1.09)^{t-1} * (1.09)^{-1}.
Therefore,
Balance = 4000 * (1.09)^{44} * sum_{k=0}^{44} (1.03/1.09)^{k}, where k = t-1.

This is a finite geometric series. Let r = 1.03/1.09. Then r ≈ 0.9450. The sum of 45 terms is:
Sum = [1 - r^{45}] / [1 - r].
Compute r^{45} by taking natural logs: ln(r) ≈ ln(1.03) - ln(1.09) ≈ 0.02956 - 0.08618 ≈ -0.05662. Then 45 * ln(r) ≈ -2.5479, so r^{45} ≈ e^{-2.5479} ≈ 0.0785 (approximately). Thus, 1 - r^{45} ≈ 0.9215.
Next, 1 - r ≈ 1 - 0.9450 ≈ 0.0550.
So Sum ≈ 0.9215 / 0.0550 ≈ 16.75.

Now compute the growth factor at the end: (1.09)^{44}. Using ln(1.09) ≈ 0.08618, we have 44 * ln(1.09) ≈ 3.794, so (1.09)^{44} ≈ e^{3.794} ≈ 44.4.

Putting it all together:
Balance ≈ 4000 * 44.4 * 16.75.
First multiply 44.4 * 16.75 ≈ 743. 
Then 743 * 4000 ≈ 2,972,000.

Since we rounded intermediate steps, the exact figure will differ slightly. The instructions require rounding to the nearest dollar, so you would present the final numerical value rounded accordingly (e.g., around $2,970,000–$2,980,000 depending on precise intermediate calculations). The key steps are:
- express the balance as a sum of growing contributions each compounded for remaining years
- convert to a geometric-series form
- evaluate the geometric sum and the annual growth/discount factors
- multiply through to obtain the final balance, then round to the nearest dollar.

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