Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 2 / c / i

Table of contents
- Solution i

Solution

n = 3

is

\frac{1}{3 ( 3 - 2 )} = \frac{1}{3} = \frac{3}{4} - \frac{5}{12} .

Assume the displayed formula holds at some

n \geq 3

. Adding the next term yields

i = 3 \sum n + 1 \frac{1}{i ( i - 2 )} = \frac{3}{4} - \frac{2 n - 1}{2 n ( n - 1 )} + \frac{1}{( n + 1 ) ( n - 1 )} = \frac{3}{4} - \frac{( 2 n - 1 ) ( n + 1 ) - 2 n}{2 n ( n - 1 ) ( n + 1 )} = \frac{3}{4} - \frac{2 n + 1}{2 n ( n + 1 )} .

The last step uses

(2 n - 1) (n + 1) - 2 n = (2 n + 1) (n - 1)

. This is the required formula at

n + 1

, completing the proof by mathematical induction. The identity holds for all $n \geq 3$ .

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