 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability/2/b/ii/solution

Use $G_{1} = 0$ and $G_{n + 1} = n G_{n} + n - 1$ for $n \geq 1$ . Indeed,

G_{n + 1} = n! - 1 = n ((n - 1)! - 1) + n - 1 = n G_{n} + n - 1.

The initial value is

0! - 1 = 0

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