1. **Problem Statement:** (i) Prove the sum of the first $n$ terms $S_n$ of a geometric series with first term $a$ and common ratio $r$ is given by $$S_n = \frac{a(1 - r^n)}{1 - r}$$ (ii) A liquid evaporates reducing volume by 8% each year from an initial 150 litres. (a) Show the volume after 6 years is approximately 91 litres. (b) Calculate total volume in 40 barrels after 40 years, each filled at the start of each year. 2. **Formula for sum of geometric series:** The sum of the first $n$ terms of a geometric series is $$S_n = a + ar + ar^2 + \cdots + ar^{n-1}$$ Multiplying both sides by $r$: $$rS_n = ar + ar^2 + ar^3 + \cdots + ar^n$$ Subtracting these equations: $$S_n - rS_n = a - ar^n$$ Factoring: $$S_n(1 - r) = a(1 - r^n)$$ Dividing both sides by $(1-r)$ (assuming $r \neq 1$): $$S_n = \frac{a(1 - r^n)}{1 - r}$$ 3. **Explanation:** This formula sums all terms by eliminating intermediate terms through subtraction, a key technique in geometric series. 4. **Part (ii)(a) - Volume after 6 years:** Initial volume $V_0 = 150$ litres. Each year volume reduces by 8%, so common ratio $r = 1 - 0.08 = 0.92$. Volume after 6 years: $$V_6 = V_0 \times r^6 = 150 \times 0.92^6$$ Calculate $0.92^6$: $$0.92^6 \approx 0.6065$$ So, $$V_6 \approx 150 \times 0.6065 = 90.975$$ Approximately 91 litres. 5. **Part (ii)(b) - Total volume in 40 barrels after 40 years:** Each year a new barrel is filled with 150 litres. At the end of 40 years, the first barrel has evaporated for 40 years, the second for 39 years, ..., the 40th barrel for 1 year. Total volume $T$ is sum of volumes in each barrel: $$T = 150(0.92^{40} + 0.92^{39} + \cdots + 0.92^1)$$ This is a geometric series with first term $a = 150 \times 0.92$ and ratio $r = 0.92$, number of terms $n=40$. Using sum formula: $$T = 150 \times 0.92 \times \frac{1 - 0.92^{40}}{1 - 0.92}$$ Calculate denominator: $$1 - 0.92 = 0.08$$ Calculate $0.92^{40} \approx 0.0397$ Calculate numerator: $$1 - 0.0397 = 0.9603$$ Calculate sum: $$T = 150 \times 0.92 \times \frac{0.9603}{0.08} = 150 \times 0.92 \times 12.00375$$ Calculate: $$150 \times 0.92 = 138$$ $$138 \times 12.00375 \approx 1656.5$$ Rounded to nearest litre: $$\boxed{1657}$$ litres. **Final answers:** (i) $$S_n = \frac{a(1 - r^n)}{1 - r}$$ (ii)(a) Approximately 91 litres after 6 years. (ii)(b) Approximately 1657 litres in total after 40 years.