 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/7/c/i/solution

- n - 1

C 2^{n}

. The initial value gives

C - 1 = 1

, so

u_{n} = 2^{n + 1} - n - 1.

Substitution verifies

u_{n + 1} - 2 u_{n} = n

, and the recurrence plus

u_{0}

determines all later terms uniquely.

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