Loans and Annuities

During each period the balance grows by its interest, then the repayment is subtracted: \(B_0 = P\) and \(B_{j+1} = B_j(1+i) - R\) for every whole number \(j \geq 0\). The first balances are $$\begin{aligned}[t] B_1 &= P(1+i) - R\\ B_2 &= B_1(1+i) - R = P(1+i)^2 - R(1+i) - R\\ B_3 &= B_2(1+i) - R = P(1+i)^3 - R(1+i)^2 - R(1+i) - R, \end{aligned}$$ which suggests, after \(k\) periods, a bracket that is the sum of the first \(k\) terms of the geometric sequence with first term \(1\) and ratio \(1+i \neq 1\): $$\begin{aligned}[t] B_k &= P(1+i)^k - R\left[1 + (1+i) + \cdots + (1+i)^{k-1}\right]\\ &= P(1+i)^k - R \times \frac{1 - (1+i)^k}{1 - (1+i)} && \text{(sum of a geometric sequence)}\\ &= P(1+i)^k - R \times \frac{(1+i)^k - 1}{i}. \end{aligned}$$ This formula holds for every whole number \(k\), by induction. For \(k = 0\) it gives \(B_0 = P - R \times 0 = P\). If it holds for \(B_j\), then $$\begin{aligned}[t] B_{j+1} &= B_j(1+i) - R\\ &= P(1+i)^{j+1} - R \times \frac{(1+i)^{j+1} - (1+i)}{i} - R\\ &= P(1+i)^{j+1} - R \times \frac{(1+i)^{j+1} - (1+i) + i}{i}\\ &= P(1+i)^{j+1} - R \times \frac{(1+i)^{j+1} - 1}{i}, \end{aligned}$$ which is the formula for \(B_{j+1}\). The loan is repaid after \(n\) repayments when \(B_n = 0\): $$\begin{aligned}[t] P(1+i)^n &= R \times \frac{(1+i)^n - 1}{i}\\ R &= \frac{P \times i \times (1+i)^n}{(1+i)^n - 1} && \text{(multiplying both sides by \(\tfrac{i}{(1+i)^n - 1} > 0\), since \(i > 0\))}. \end{aligned}$$