 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/discrete-mathematics-continuous-mathematics-and-probability/2/a/ii/solution

For

n = 1

, both sides are

1/3

. Assuming the formula through

n

, adding the next term gives

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

The last expression is the required formula at

n + 1

\frac{1}{( 2 i - 1 ) ( 2 i + 1 )} = \frac{1}{2} (\frac{1}{2 i - 1} - \frac{1}{2 i + 1})

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