This is a question from TJC Promotion 2017 Question 10. Thank you Mr. Wee for sharing.

Mr. Scrimp started a savings account which pays compound interest at a rate of r% per year on the last day of each year.

He made an initial deposit of $ x on 1 January 2000. From 1 January 2001 onwards, he makes a deposit of $ x at the start of each year.

(i) Show that the total amount in the savings account at the end of the n ^{th} year is $ \frac{k}{k-1} (k^n - 1)x , where k = \frac{100 + r}{100}.

(ii) At the end of the year 2019, Mr. Scrimp has a total amount of $ 22x in the savings account. Find the value of r, giving your answer correct to one decimal place.

Assume that the last deposit is made on 1 January 2019 and that the total amount in the savings account is $50000 on 1 January 2020.

For a period of N years, where 1 \le N \le 20, Mr. Scrimp can either continue to keep this amount in the savings account to earn interest or invest this amount in a financial product. The financial product pays an upfront sign-up bonus of $2000 and a year-end profit of $200 in the first year. At the end of each subsequent year, the financial product pays $20 more profit that in the previous year.

(iii) Find the total amount Mr. Scrimp will have at the end of N years if he invests in the financial product.

(iv) Using the value of r found in (ii), find the maximum number of years Mr. Scrimp should invest in the financial product for it to be more profitable than keeping the money in the savings account.

(i)
(1 + \frac{r}{100})x  + (1 + \frac{r}{100})^2 x + (1 + \frac{r}{100})^3 x + \ldots + (1 + \frac{r}{100})^n x

= \frac{(1 + \frac{r}{100})x[(1 + \frac{r}{100})^n - 1]}{(1 + \frac{r}{100}) - 1}

= \frac{(\frac{100 + r}{100})x[(\frac{100 + r}{100})^n - 1]}{(\frac{100 + r}{100}) - 1}

Let k = \frac{100 + r}{100}

\Rightarrow \frac{k x[k^n - 1]}{k - 1}

Total Amount = \$ [\frac{k}{k-1}(k^n - 1 ) x ]

(ii)

\frac{k}{k-1}(k^{20} - 1 ) x  = 22 x

\frac{k}{k-1}(k^{20} - 1 )  = 22

Using GC, k = 1.0089905

\Rightarrow \frac{100 + r}{100} = 1.0089905

\therefore r = 0.89905 \approx 0.9 (1 decimal place)

(iii)

Total amount = 50000 + 2000 + \frac{N}{2}[2(200) + (N-1)(2)]

= 52000 + N(200 + N-1)

= 52000 + 199N + N^2

(iv)

With the savings account, he has $ [50000(1.009)^N] at the end of N years.

With the financial product, he has $ [52000 + 199N + N^2] at the end of N years.

For it to be more profitable,

[52000 + 199N + N^2] > [50000(1.009)^N]

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