 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/i/solution

n = 2

, the product is

1 - 1/4 = 3/4 = (2 + 1) / (2 \cdot 2)

. Assume the identity holds for some

n \geq 2

. Multiplying by the next factor gives

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

This is the desired expression with

n

replaced by

n + 1

, so mathematical induction proves the identity for every

n \geq 2

1 - i^{- 2} = (i - 1) (i + 1) / i^{2}

also explains the cancellation as a telescoping product.

 New to topics? Read the docs here!