 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/7/b/solution

The initial term at

n = - 1

is three, and the common ratio is

1/3

. The convergent geometric series therefore sums to

n = - 1 \sum \infty (\frac{1}{3})^{n} = \frac{3}{1 - 1/3} = \frac{9}{2} .

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