 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iii/4/b/v/solution

n /2

. Subtracting it gives

r_{k + 1} - n /2 = (1 - 2/ n) (r_{k} - n /2) .

Consequently

r_{k} = \frac{n}{2} + \frac{n}{2} (1 - \frac{2}{n})^{k}, k \geq 0.

For

n = 2

, interpret the formula at

k = 0

r_{0} = 2

, while

r_{k} = 1

for

k \geq 1

. For every

n \geq 2

r_{k} \to n /2

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