 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2022/discrete-mathematics-continuous-mathematics-and-probability/discrete-mathematics/2/b/solution

n \geq 0

. For

n = 0

, the sum is

0! 0 = 0 = 1! - 1

. Suppose it is

(n + 1)! - 1

n

. Then

\sum_{i = 0}^{n + 1} i! i = (n + 1)! - 1 + (n + 1)! (n + 1) = (n + 2)! - 1.

The induction is complete, proving $\sum_{i = 0}^{n} i! i = (n + 1)! - 1$ . The factorial identity

i! i = (i + 1)! - i!

also explains the cancellation directly.

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