 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/8/solution

Substitute

r = - x^{2}

into the geometric series, without taking any derivatives:

\frac{1}{1 + x ^{2}} = k = 0 \sum \infty (- 1)^{k} x^{2 k} = 1 - x^{2} + x^{4} - \dots, ∣ x ∣ < 1.

Thus the coefficient of

x^{n}

(- 1)^{n /2}

when

n

is even and zero when

n

is odd. If nonzero terms are numbered from one, the

n

th term is

(- 1)^{n - 1} x^{2 n - 2}

 New to topics? Read the docs here!