 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/vi/solution

The even-power formula reduces every summand to a scalar multiple of

I

\sum_{n = 1}^{\infty} \frac{A ^{4 n} - 2 A ^{2 n}}{n} = [\sum_{n = 1}^{\infty} \frac{( 1/16 ) ^{n}}{n} - 2 \sum_{n = 1}^{\infty} \frac{( - 1/4 ) ^{n}}{n}] I .

\sum_{n \geq 1} z^{n} / n = - lo g (1 - z)

for

∣ z ∣ < 1

to obtain

[- lo g \frac{15}{16} + 2 lo g \frac{5}{4}] I = lo g \frac{5}{3} I .

Thus the off-diagonal entries vanish and both diagonal entries have the required value.

 New to topics? Read the docs here!