 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/7/a/solution

There are

100

terms in this arithmetic series, starting at

4

and ending at

202

. Pairing the first and last terms, the second and penultimate terms, and so on, gives

n = 2 \sum 101 2 n = \frac{100}{2} (4 + 202) = 10300.

The summand in the PDF is

2 n

, not

2^{n}

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