**Topic**

Assume that in the first year, Ms. Hanh deposits in bank A (VND) with interest rate r%/year. At the end of the first year, Ms. Hanh did not withdraw money and deposited more A (VND). At the end of the second year, Ms. Hanh also did not withdraw money and deposited A (dong) again. Continue like this for years to come. Prove the amount of both capital and interest that Ms. Hanh has *Okay* after n (years) is \({T_n} = \frac{{A(100 + r)}}{r}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} – 1} \right]\) (VND), if during this period the interest rate remains unchanged.

**Detailed explanation**

We prove that “The amount of both capital and interest that Ms. Hanh has after n (years) is \({T_n} = \frac{{A(100 + r)}}{r}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^n} – 1} \right]\) (dong)” by the induction method.

Step 1: When \(n = 1\) we have

The amount of both capital and interest that Ms. Hanh has after 1 year is: \(A + r\% .A = A.\left( {1 + \frac{r}{{100}}} \right) = \ frac{{A(100 + r)}}{{100}}\)(bronze)

And \({T_1} = \frac{{A(100 + r)}}{r}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^1} – 1} \right] = \frac{{A(100 + r)}}{r}.\frac{r}{{100}} = \frac{{A(100 + r)}}{{100}}\)(bronze)

So the statement is true for \(n = 1\)

Step 2: With k being an arbitrary positive integer whose proposition is true, we have to prove the proposition to be true for k + 1, that is:

“The amount of both capital and interest that Ms. Hanh has after \(k + 1\) years is: \({T_{k + 1}} = \frac{{A(100 + r)}}{r}\ left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^{k + 1}} – 1} \right]\) (copper)”

Indeed, by the assumption of induction we have:

The amount of both capital and interest that Ms. Hanh has after \(k\) years is: \({T_k} = \frac{{A(100 + r)}}{r}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^k} – 1} \right]\) (copper)

She didn’t withdraw, but sent another A

=> The principal amount after \(k + 1\) years is: \({T_k} + A\)(VND)

=> The profit after \(k + 1\) years is: \(\left( {{T_k} + A} \right).r\% \)(VND)

The amount of both capital and interest that Ms. Hanh has after \(k + 1\) years is:

\(\begin{array}{l}{T_k} + A + \left( {{T_k} + A} \right).r\% = \left( {{T_k} + A} \right).(1 + r\% ) = \left( {{T_k} + A} \right)\left( {1 + \frac{r}{{100}}} \right)\\ = \left\{ {\frac{ {A(100 + r)}}{r}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^k} – 1} \right] + A} \right\}.\left( {1 + \frac{r}{{100}}} \right)\\ = \frac{{A(100 + r)}}{r}\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^k} – 1} \right].\left( {1 + \frac{r}{{100}}} \right) + A.\left( {1 + \frac{r}{{100}}} \right)\\ = \frac{ {A(100 + r)}}{r}.\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^{k + 1}} – \left( {1 + \frac{r}{{100}}} \right)} \right] + A.\left( {\frac{{100 + r}}{{100}}} \right)\\ = \frac{{A(100 + r)}}{r}.\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^{k + 1}} – \left( {1 + \frac{r}{{100}}} \right)} \right] + A.\left( {\frac{{100 + r}}{r}} \right).\frac{r}{{100}}\\ = \frac{{A(100 + r)}}{ r}.\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^{k + 1}} – \left( {1 + \frac{r}{{100}}} \right) + \frac{r}{{100}}} \right]\\ = \frac{{A(100 + r)}}{r}.\left[ {{{\left( {1 + \frac{r}{{100}}} \right)}^{k + 1}} – 1} \right]\end{array}\)

So the statement is true for k+1. Thus, according to the principle of mathematical induction, the statement is true for all \(n \in \mathbb{N}*\).