 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ia/paper-4/6d/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

(1 + t)^{n} = \sum_{k = 0}^{n} (k n) t^{k}

n

1 + t

and collection of the coefficient of

t^{k}

0

1

gives

k = 0 \sum n \frac{1}{k + 1} (k n) = \int_{0}^{1} (1 + t)^{n} d t = \frac{2 ^{n + 1} - 1}{n + 1} .

Replacing

t

- t

gives

k = 0 \sum n \frac{( - 1 ) ^{k}}{k + 1} (k n) = \int_{0}^{1} (1 - t)^{n} d t = \frac{1}{n + 1} .

Finally,

\sum_{k = 1}^{n} \frac{( - 1 ) ^{k + 1}}{k} (k n) = \int_{0}^{1} \frac{1 - ( 1 - x ) ^{n}}{x} d x .

y = 1 - x

and using the finite geometric series,

\frac{1 - y ^{n}}{1 - y} = 1 + y + \dots + y^{n - 1},

turns this into

j = 0 \sum n - 1 \int_{0}^{1} y^{j} d y = 1 + \frac{1}{2} + \dots + \frac{1}{n} .

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