Sarah 是一名新加坡公民，于 2021 年 8 月在新加坡国立大学 (NUS) 开始了为期 4 年的本科计算机课程。在扣除教育部学费津贴 (MOE Tuition Grant) 后，每年的学费为 $8,250。Sarah 申请了覆盖其学费 $90\%$ 的学费贷款 (TFL)。该贷款在她的学习期间是免息的。 Sarah 决定从 2021 年 8 月起，每月存下 $x$ 。此后每个月该金额增加 $2。（Sarah 在 2021 年 8 月存下 $x$ ，在 2021 年 9 月存下 $(x+2)$ ，…）。利用存下的这笔钱，她在 2025 年 7 月底偿还了其 TFL 总额的五分之一。在 Sarah 毕业后，TFL 的利息开始生效，固定月利率为 $0.4\%$ ，按每月末的未还余额计算。利息于 2025 年 8 月 31 日首次计息。 Sarah 计划从 2025 年 9 月 1 日开始，在每月的 1 日向银行月还款 $y$ 。

(a)
求 $x$ 的值。
(b)
求 Sarah 进行第一次月还款后的未还贷款金额。
(c)
证明在 Sarah 进行第 $n$ 次月还款后，未还贷款金额为

$\displaystyle 23760(1.004)^n - 250y(1.004^n - 1)。$
(d)
Sarah 希望在 3 年底前付清剩余的贷款。求 $y$ 的值。
(e)
Sarah 决定采用不同的还款策略。在前 24 个月内，她每月还款 $500。在这 24 个月之后，她将每月还款额加倍至 $1000。

确定 Sarah 结清贷款的月份和年份，并计算她最后一次还款的金额。

等差数列与等差级数递推关系等比数列与等比级数解方程数列与函数的关系通项公式 English

先不看答案，让 AI 诊断你卡在哪里？

它不会直接告诉你答案，而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

12(a)

1. Calculate the total loan amount and the target amount to be repaid.

$\text{Total tuition fee for 4 years} = \char36{}8250 \times 4 = \char36{}33000$

$\text{Total TFL} = 90\% \text{ of } \char36{}33000 = 0.9 \times \char36{}33000 = \char36{}29700$

$\text{Amount repaid} = \frac{1}{5} \times \char36{}29700 = \char36{}5940$

2. Express the total amount set aside over the 4 years in terms of $x$ using the arithmetic series sum formula.

The duration from August 2021 to July 2025 is exactly 4 years, which is $48$ months.

The amounts set aside form an arithmetic progression with first term $a = x$ , common difference $d = 2$ , and number of terms $n = 48$ .

$S_{48} = \frac{n}{2} \left[ 2a + (n - 1)d \right]$
$S_{48} = \frac{48}{2} \left[ 2x + (48 - 1)(2) \right]$
$S_{48} = 24(2x + 94)$
$S_{48} = 48x + 2256$

3. Formulate an equation by equating the total savings to the amount repaid, and solve for $x$.

$48x + 2256 = 5940$
$48x = 3684$
$x = 76.75$

答案

76.75

12(b)

1. Calculate the total Tuition Fee Loan (TFL) and the outstanding loan principal before interest begins.

$\text{Total tuition fees} = 4 \times 8250 = 33000$
$\text{Total TFL taken} = 0.9 \times 33000 = 29700$

Since Sarah repaid $\frac{1}{5}$ of her total TFL at the end of July 2025, the outstanding amount is:
$\text{Outstanding loan (end July 2025)} = \frac{4}{5} \times 29700 = 23760$

2. Apply the first interest charge and subtract the first monthly repayment.

Interest is first charged on 31st August 2025 at $0.4\%$ per month.
$\text{Balance on 31st August 2025} = 23760 \times \left(1 + \frac{0.4}{100}\right)$
$= 23760 \times 1.004 = 23855.04$

After her first monthly repayment of $y$ on 1st September 2025:
$\text{Outstanding loan} = 23855.04 - y$

答案

$23855.04 - y$

12(c)

1. Calculate the outstanding loan for the first few months to establish a pattern.

\begin{aligned} \text{Initial loan amount} &= 4 \times 8250 \times 0.9 \times \left(1 - \frac{1}{5}\right) = 23760 \\ \text{After 1st repayment} &= 23760(1.004) - y \\ \text{After 2nd repayment} &= [23760(1.004) - y](1.004) - y \\ &= 23760(1.004)^2 - y(1.004) - y \\ \text{After 3rd repayment} &= [23760(1.004)^2 - y(1.004) - y](1.004) - y \\ &= 23760(1.004)^3 - y(1.004)^2 - y(1.004) - y \end{aligned}

2. Generalise the outstanding loan expression for $n$ months.

\text{After } n\text{th repayment} = 23760(1.004)^n - y(1.004^{n-1} + 1.004^{n-2} + \dots + 1.004 + 1)

3. Apply the sum formula for a finite geometric series and substitute to show the final expression.

\begin{aligned} 1.004^{n-1} + 1.004^{n-2} + \dots + 1.004 + 1 &= \frac{1(1.004^n - 1)}{1.004 - 1} \\ &= \frac{1.004^n - 1}{0.004} \\ &= 250(1.004^n - 1) \\ \text{Outstanding loan} &= 23760(1.004)^n - y[250(1.004^n - 1)] \\ &= 23760(1.004)^n - 250y(1.004^n - 1) \quad \text{(shown)} \end{aligned}

答案

$23760(1.004)^n - 250y(1.004^n - 1)$

12(d)

1. Set up the equation for the loan to be paid off in 3 years.

Since the loan is paid off in 3 years, the number of monthly repayments is $n = 3 \times 12 = 36$ .
Equating the expression from part (c) to zero:
$23760(1.004)^{36} - 250y(1.004^{36} - 1) = 0$

2. Solve for $y$.

$250y(1.004^{36} - 1) = 23760(1.004)^{36}$
$250y \approx \frac{27432.0189}{0.154546}$
$y = \frac{23760(1.004)^{36}}{250(1.004^{36} - 1)}$
$y \approx 710.0007$
$y = 710.00 \text{ (to 2 d.p. for money)}$

答案

$710.00$

12(e)

1. Calculate the outstanding loan after 24 months of $500 repayments.

Using the formula from (c) with $y = 500$ and $n = 24$ :
$\text{Outstanding loan} = 23760(1.004)^{24} - 250(500)(1.004^{24} - 1)$
$= 13576.5154$

2. Set up an equation to find the additional number of months $m$ required to clear the loan with $1000 repayments.

$13576.5154(1.004)^m - 250(1000)(1.004^m - 1) = 0$
$13576.5154(1.004)^m - 250000(1.004^m) + 250000 = 0$
$236423.4846(1.004)^m = 250000$
$(1.004)^m = \frac{250000}{236423.4846} \approx 1.05742$
$m = \frac{\ln 1.05742}{\ln 1.004} \approx 13.987$
Since $m$ must be a whole number of repayments, 14 additional months are needed.

3. Determine the calendar month and year when the loan is cleared.

Total number of months $= 24 + 14 = 38$ .
The 1st repayment is in September 2025. The 38th repayment occurs 37 months later.
37 months is 3 years and 1 month.
September 2025 $+ 3$ years $=$ September 2028.
September 2028 $+ 1$ month $=$ October 2028.

4. Calculate the amount of the final repayment in the 38th month.

The final repayment is made in the 14th additional month (month 38 overall). First, find the outstanding loan after 13 additional payments (the 37th payment overall):
$A_{37} = 13576.5154(1.004)^{13} - 250000(1.004^{13} - 1)$
$A_{37} \approx 981.6602$
The final repayment covers this remaining balance plus 1 month of interest:
$\text{Final repayment} = 981.6602 \times 1.004 \approx 985.59$

答案

October 2028, $\char36{}985.59$

AI 分析，可能出错

用 AI 批改，圈出写错的那一步，并自动将你的薄弱点存入错题本，形成你的错因库。