 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/6/solution

r = 1/ 2

. Since

r^{10} = 1/32

\sum_{n = 1}^{10} 2^{- n /2} = \frac{r ( 1 - r ^{10} )}{1 - r} = \frac{31}{32} \frac{1}{2 - 1} = \frac{31 + 31 2}{32} .

Thus

p = 31, q = 31

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