 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-iii/5/b/iii/solution

16 r^{2} - 8 r - 3 = (4 r - 3) (4 r + 1) = 0.

Its distinct roots are

3/4

and

- 1/4

, so

p_{n} = A (3/4)^{n} + B (- 1/4)^{n}

p_{0} = 1

and

p_{1} = 1/2

A + B = 1

and

3 A - B = 2

. Hence

A = 3/4

B = 1/4

, and

p_{n} = \frac{3}{4} (\frac{3}{4})^{n} + \frac{1}{4} (- \frac{1}{4})^{n} = \frac{3 ^{n + 1} + ( - 1 ) ^{n}}{4 ^{n + 1}}, n \geq 0.

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