(i)     Find an expression for \({S_1}\) and show that

\[{S_2} = P{\left( {1 + \frac{I}{{100}}} \right)^2} - R\left( {1 + \left( {1 + \frac{I}{{100}}} \right)} \right).\]

(ii)     Determine a similar expression for \({S_n}\) . Hence show that

\[{S_n} = P{\left( {1 + \frac{I}{{100}}} \right)^n} - \frac{{100R}}{I}\left( {{{\left( {1 + \frac{I}{{100}}} \right)}^n} - 1} \right)\]

Sue borrows $5000 at a monthly interest rate of 1 % and plans to repay the loan in 5 years (i.e. 60 months).

(i)     Calculate the required monthly repayment, giving your answer correct to two decimal places.

(ii)     After 20 months, she inherits some money and she decides to repay the loan completely at that time. How much will she have to repay, giving your answer correct to the nearest $?

(i)     \({S_1} = P\left( {1 + \frac{I}{{100}}} \right) - R\)     A1

\({S_2} = P{\left( {1 + \frac{I}{{100}}} \right)^2} - R\left( {1 + \frac{I}{{100}}} \right) - R\)     M1A1

\( = P{\left( {1 + \frac{I}{{100}}} \right)^2} - R\left( {1 + \left( {1 + \frac{I}{{100}}} \right)} \right)\)     AG

(ii)     extending this,

\({S_n} = P{\left( {1 + \frac{I}{{100}}} \right)^n} - R\left( {1 + \left( {1 + \frac{I}{{100}}} \right) + \ldots + {{\left( {1 + \frac{I}{{100}}} \right)}^{n - 1}}} \right)\)     M1A1

\( = P{\left( {1 + \frac{I}{{100}}} \right)^n} - \frac{{R\left( {{{\left( {1 + \frac{I}{{100}}} \right)}^n} - 1} \right)}}{{\frac{I}{{100}}}}\)     M1A1

\( = P{\left( {1 + \frac{I}{{100}}} \right)^n} - \frac{{100R}}{I}\left( {{{\left( {1 + \frac{I}{{100}}} \right)}^n} - 1} \right)\)     AG

[7 marks]

(i)     putting \({S_{60}} = 0,{\text{ }}P = 5000,{\text{ }}I = 1\)     M1

\(5000 \times {1.01^{60}} = 100R{\text{ }}({1.01^{60}} - 1)\)     A1

\(R = (\$ )111.22\)     A1

(ii)     \(n = 20,{\text{ }}P = 5000,{\text{ }}I = 1,{\text{ }}R = 111.22\)     M1

\({S_{20}} = 5000 \times {1.01^{20}} - 100 \times 111.22({1.01^{20}} - 1)\)     A1

\( = (\$ )3652\)     A1

which is the outstanding amount

[6 marks]