 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/12/b/v/solution

Repeatedly using the quadratic identity gives

A^{2 n} = (- \frac{1}{4})^{n} I, A^{2 n + 1} = (- \frac{1}{4})^{n} A .

In entries these are

(- 1)^{n} 4^{- n} (1001)

and

(- 1)^{n} 2^{- (2 n + 1)} (0 - 1 10)

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