A-Level Sequences and series: worked solution

Question

Show that after her nth monthly repayment the outstanding loan is 23760(1.004)ⁿ − 250y(1.004ⁿ − 1).

Let uₙ be the amount still owed immediately after the nth monthly repayment. Each month the outstanding amount first grows by the monthly interest rate 0.4%, i.e. it is multiplied by 1.004, and then the repayment of y dollars is subtracted: uₙ = 1.004uₙ₋₁ − y. Build up the first few terms from the initial loan of 23760 dollars, so that the pattern is visible: u₁ = 23760(1.004) − y u₂ = 1.004u₁ − y = 23760(1.004)² − y(1.004) − y u₃ = 1.004u₂ − y = 23760(1.004)³ − y(1.004)² − y(1.004) − y Each repayment made k months ago has itself accrued interest for k months, so after n repayments uₙ = 23760(1.004)ⁿ − y[1 + 1.004 + 1.004² + ... + 1.004ⁿ⁻¹]. The bracket is a geometric series with first term 1, common ratio r = 1.004 and n terms, so its sum is (1.004ⁿ − 1)/(1.004 − 1) = (1.004ⁿ − 1)/0.004. Since 1/0.004 = 250, this is 250(1.004ⁿ − 1). Therefore uₙ = 23760(1.004)ⁿ − 250y(1.004ⁿ − 1), as required.