Answer
a. \[
A_{k} = 1.0025\,A_{k-1} + 200, \quad A_{0} = 1000.
\]
b. \[
A_{k}
= 1000\,(1.0025)^k
\;+\; 200\,\frac{(1.0025)^k - 1}{\,1.0025 - 1\,}.
\]
c. **Induction Proof:** Standard for linear recurrences (shown above).
d. \(\displaystyle A_{240} \approx \$66{,}610.\)
\(\displaystyle A_{480} \approx \$185{,}680.\)
e. **Time to reach \$10,000:** About **42 months** (3.5 years). The exact crossing occurs near \(n \approx 42.2\) months, so in the 43rd month.
Work Step by Step
Below is a step‐by‐step solution outline. We let
- \(i = 3\%\) per year be the nominal annual interest rate.
- Interest is compounded **monthly**, so the monthly interest rate is \(\tfrac{i}{12} = \tfrac{0.03}{12} = 0.0025\) (i.e.\ 0.25% per month).
- Let \(r = 1 + \tfrac{i}{12} = 1.0025.\)
- Each month, right **after** interest is added, the person deposits an additional \$200.
- \(A_n\) is the amount in the account at the end of \(n\) months.
- The initial deposit (at month 0) is \$1{,}000, so \(A_0 = 1000.\)
We want:
a. A recurrence relation for \(A_{k}\) in terms of \(A_{k-1}\).
b. An explicit (closed‐form) formula for \(A_n\).
c. A brief induction proof.
d. The value of the account after 20 years and after 40 years.
e. The time (in years) at which the account first reaches \$10,000.
---
## (a) Recurrence Relation
At the end of each month \(k\):
1. The previous amount \(A_{k-1}\) has earned one month of interest, so it becomes \(r \, A_{k-1}\).
2. Then \$200 is deposited.
Hence the recurrence is
\[
\boxed{A_{k} \;=\; r \, A_{k-1} \;+\; 200, \quad\text{for } k \ge 1,\quad\text{with } A_{0} = 1000.}
\]
where \(r = 1.0025.\)
---
## (b) Iteration to Find an Explicit Formula
We can “unfold” the recurrence:
\[
\begin{aligned}
A_{k}
&= r \, A_{k-1} + 200 \\
&= r\bigl(r \, A_{k-2} + 200\bigr) + 200
= r^2 A_{k-2} + r \cdot 200 + 200 \\
&= r^3 A_{k-3} + r^2 \cdot 200 + r \cdot 200 + 200 \\
& \;\;\vdots \\
&= r^k A_{0} \;+\; 200 \bigl(r^{k-1} + r^{k-2} + \cdots + r + 1\bigr).
\end{aligned}
\]
Since \(A_0 = 1000\), we get
\[
A_{k}
= 1000\,r^{k}
\;+\; 200 \sum_{j=0}^{k-1} r^{\,j}.
\]
The finite geometric sum is
\[
\sum_{j=0}^{k-1} r^j
= \frac{r^k - 1}{r - 1}.
\]
Thus,
\[
\boxed{
A_{k}
= 1000 \, r^{k}
\;+\; 200 \,\frac{r^{k} - 1}{r - 1},
\quad\text{where } r = 1.0025.
}
\]
That is our explicit formula.
---
## (c) Proof by Mathematical Induction (Sketch)
**Claim:**
\[
A_{k}
= 1000 \, r^{k}
\;+\; 200 \,\frac{r^{k} - 1}{r - 1}.
\]
- **Base Case** \((k=0)\).
The right‐hand side for \(k=0\) gives
\[
1000\,r^0 + 200 \cdot \frac{r^0 - 1}{r - 1}
= 1000 \cdot 1 + 200 \cdot \frac{1 - 1}{r - 1}
= 1000 + 0 = 1000.
\]
This matches \(A_0 = 1000\).
- **Inductive Step.**
Assume it holds for some \(k\). Then for \(k+1\):
\[
A_{k+1}
\;=\; r \, A_k + 200
\;=\; r \Bigl[1000\,r^k + 200\,\frac{r^k - 1}{r-1}\Bigr] + 200.
\]
Factor out \(r^{k+1}\) and combine terms. One finds it simplifies exactly to
\[
1000\,r^{k+1} \;+\; 200\,\frac{r^{k+1} - 1}{r-1},
\]
completing the induction.
---
## (d) Value After 20 and 40 Years
- 1 year = 12 months, so
- 20 years = 240 months,
- 40 years = 480 months.
Hence we compute
\[
A_{240}
= 1000\,r^{240}
\;+\; 200\,\frac{r^{240} - 1}{r-1},
\quad
A_{480}
= 1000\,r^{480}
\;+\; 200\,\frac{r^{480} - 1}{r-1},
\]
where \(r = 1.0025\).
**Numerical Approximation** (roughly):
- \(r^{240} \approx (1.0025)^{240} \approx 1.81\) to 1.82.
- \(r^{480} \approx (1.0025)^{480} \approx (r^{240})^2 \approx 1.81^2 \approx 3.28\) to 3.30.
Putting these into the formula:
1. **After 20 years (240 months):**
\[
A_{240}
\approx 1000 \times 1.81
\;+\; 200 \times \frac{1.81 - 1}{0.0025}.
\]
That is roughly
\[
1810
\;+\; 200 \times \frac{0.81}{0.0025}
\;=\; 1810
\;+\; 200 \times 324
\;=\; 1810 + 64{,}800
\;=\; 66{,}610\text{ dollars (approx)}.
\]
2. **After 40 years (480 months):**
\[
A_{480}
\approx 1000 \times 3.28
\;+\; 200 \times \frac{3.28 - 1}{0.0025}
\]
\[
\approx 3280
\;+\; 200 \times \frac{2.28}{0.0025}
\;=\; 3280 + 200 \times 912
\;=\; 3280 + 182{,}400
\;=\; 185{,}680\text{ dollars (approx)}.
\]
Exact values can be found by more precise computation, but these illustrate the magnitudes.
---
## (e) When Does the Account Reach \$10,000?
We want the smallest integer \(n\) such that \(A_n \ge 10{,}000.\)
Set up the equation
\[
A_{n}
= 1000\,r^n + 200\,\frac{r^n - 1}{r - 1}
= 10{,}000.
\]
Combine terms. A convenient approach is to factor out \(r^n\):
\[
10{,}000
= r^n \Bigl[1000 + 200\,\tfrac{1}{r-1}\Bigr]
\;-\; 200\,\tfrac{1}{r-1},
\]
or, equivalently,
\[
10{,}000 + 200\,\frac{1}{r-1}
= r^n \Bigl[1000 + 200\,\tfrac{1}{r-1}\Bigr].
\]
But it’s often simpler just to write
\[
r^n = \frac{10{,}000(r - 1) + 200}{1000(r - 1) + 200}.
\]
With \(r - 1 = 0.0025\), we get
- Numerator = \(10{,}000 \times 0.0025 + 200 = 25 + 200 = 225.\)
- Denominator = \(1000 \times 0.0025 + 200 = 2.5 + 200 = 202.5.\)
Hence
\[
r^n
= \frac{225}{202.5}
= \frac{225}{202.5} \approx 1.111111\ldots
= \tfrac{10}{9}.
\]
Thus,
\[
n = \frac{\ln\!\bigl(\tfrac{10}{9}\bigr)}{\ln(r)}
= \frac{\ln(10/9)}{\ln(1.0025)}.
\]
Numerically,
- \(\ln(10/9) \approx 0.1053605,\)
- \(\ln(1.0025) \approx 0.002496\) (approximately),
so
\[
n \approx \frac{0.10536}{0.002496} \approx 42.2 \text{ months}.
\]
In years, that is about \(42.2 / 12 \approx 3.52\) years.
Since \(n\) must be an integer number of months, the account first exceeds \$10,000 **during the 43rd month** (which is a bit more than 3.5 years).
