 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability-resit/3/e/i/solution

T (n) = T (0) + \sum_{j = 1}^{n} j = 1 + \frac{n ( n + 1 )}{2} .

Thus

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

Indeed

T (n) / n^{2} \to 1/2

, so the sharper growth classification is

Θ (n^{2})

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