 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/13/b/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

(1 + x /3)^{- 3} = 1 - x + \frac{2}{3} x^{2} - \frac{10}{27} x^{3} + O (x^{4}), 1 - e^{x} = - x - \frac{x ^{2}}{2} - \frac{x ^{3}}{6} - \frac{x ^{4}}{24} + O (x^{5}) .

(1 - e^{x}) (1 + x /3)^{- 3} = - x + \frac{x ^{2}}{2} - \frac{x ^{3}}{3} + \frac{35 x ^{4}}{216} + O (x^{5}) .

Adding

lo g (1 + x) = x - x^{2} /2 + x^{3} /3 - x^{4} /4 + O (x^{5})

cancels the first three powers. Hence the first nonzero term is

- \frac{19}{216} x^{4} .

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