A financial institution, Future Investments Inc., has introduced a new investment scheme. The scheme pays a compound interest of 5% per annum at the end of the year, based on the amount in the account at the beginning of each year. John and Sarah are both interested in this scheme.

1. Calculate the total amount after applying the 5% interest for one year

$\text{Amount} = x(1 + 0.05) = 1.05x$

1. Establish the pattern for the amount in the account at the end of the first few years to form a series.

Amount at the end of year 1: $1.05x$
Amount at the end of year 2: $1.05(1.05x + x) = 1.05^2x + 1.05x$
Amount at the end of year 3: $1.05(1.05^2x + 1.05x + x) = 1.05^3x + 1.05^2x + 1.05x$

Total amount at the end of $n$ years:
$S_n = 1.05x + 1.05^2x + 1.05^3x + \dots + 1.05^nx$

2. Identify the parameters of the geometric progression and apply the sum formula.

The series is a geometric progression with:
First term, $a = 1.05x$
Common ratio, $r = 1.05$
Number of terms = $n$

Using the sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$:
$S_n = \frac{1.05x(1.05^n - 1)}{1.05 - 1}$

3. Simplify the expression to match the given result.

$S_n = \frac{1.05x(1.05^n - 1)}{0.05}$
$S_n = \left(\frac{1.05}{0.05}\right) x (1.05^n - 1)$
$S_n = 21x(1.05^n - 1)$ (Shown)

1. Set up the inequality for the amount at the end of $n$ years to exceed \$500 000.

Using the expression from part (a)(ii) with $x = 10\,000$:
$21(10\,000)(1.05^n - 1) > 500\,000$
$210\,000(1.05^n - 1) > 500\,000$
$1.05^n - 1 > \frac{50}{21}$
$1.05^n > \frac{71}{21}$

2. Solve for the number of years $n$ using logarithms.

$n \ln 1.05 > \ln \left(\frac{71}{21}\right)$
$n > \frac{\ln(71/21)}{\ln 1.05}$
$n > 24.967$
Since the sum is evaluated at the end of the year, $n$ must be an integer, meaning the amount evaluated at the end of the year first exceeds $500 000 when $n = 25$.

3. Check the account balance at the start of the 25th year.

We must check if the threshold was already crossed at the start of the 25th year when John makes his deposit.
Amount at the end of the 24th year:
$S_{24} = 210\,000(1.05^{24} - 1) \approx 467\,270.99$
Amount at the start of the 25th year (after adding $\char36{}10\,000$):
$467\,270.99 + 10\,000 = 477\,270.99$
Since $\char36{}477\,270.99 < \char36{}500\,000$, the amount does not yet exceed $\char36{}500\,000$ at the start of the 25th year.

4. Conclude when the amount first exceeds the threshold.

Amount at the end of the 25th year (after 5% interest is applied):
$477\,270.99 \times 1.05 \approx 501\,134.54$
This value exceeds $\char36{}500\,000$. Thus, it happens at the end of the 25th year.

1. Identify the amount deposited at the start of the $r$-th year.

Let $r$ represent the year a specific deposit is made. Sarah invests $\char36{}6000$ in the first year ($r=1$), and the amount increases by $\char36{}400$ each subsequent year. Thus, the deposit amount forms an arithmetic progression where the amount invested at the start of year $r$ is $6000 + 400(r-1)$.

2. Explain the compound interest duration for the $r$-th deposit and formulate the summation.

The deposit made at the start of year $r$ remains in the account until the end of year $n$. This corresponds to a duration of $n - r + 1$ years, which can be written as $n - (r - 1)$ years. With an annual compound interest rate of 5%, the accumulated value of this single deposit is $[6000 + 400(r-1)](1.05)^{n-(r-1)}$.

The total amount in the account at the end of $n$ years is the sum of the accumulated values of all the deposits made from year $r=1$ to year $r=n$, which gives:
$\sum_{r=1}^n [6000 + 400(r-1)](1.05)^{n-(r-1)}$

3. Determine the constants $a$ and $b$ by comparing with the given expression.

Comparing the derived summation with $\sum_{r=1}^n [a + b(r-1)](1.05)^{n-(r-1)}$, we can deduce that:
$a = 6000$
$b = 400$

1. Formulate the inequality for the total amount after $n$ years.

We require the total amount after $n$ complete years to exceed $500\,000$. Using the expression from (b)(i) with $a = 6000$ and $b = 400$, we have:
$\sum_{r=1}^n [6000 + 400(r-1)](1.05)^{n-(r-1)} > 500\,000$

2. Evaluate the sum using a Graphing Calculator to find the smallest integer $n$.

Using a Graphing Calculator to generate a table of values for the sum:
When $n = 25$, Total Amount $\approx 491\,588.37$
When $n = 26$, Total Amount $\approx 532\,967.79$

Since $532\,967.79 > 500\,000$, the amount first exceeds $\char36{}500\,000$ after 26 years.