 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/2/a/ii/solution

(2 n) + (2 n + 1) = \frac{n ( n - 1 )}{2} + \frac{n ( n + 1 )}{2} = n^{2} .

For

n = 0

n = 1

, the same equality holds with the conventional zero value when the lower index exceeds the upper index.

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