# The product of (1-1/(2^2))*(1-1/(3^2))....(1-1/(9^2))*(1-1/(10^2))=a/b then find a and b without full simplification?

Jul 17, 2018

The answer is $a = 11$ and $b = 20$

#### Explanation:

I think it's the same as

${\Pi}_{k = 2}^{10} \left(1 - \frac{1}{k} ^ 2\right)$

But I did it this way

$\left(1 - \frac{1}{2} ^ 2\right) \left(1 - \frac{1}{3} ^ 2\right) \left(1 - \frac{1}{4} ^ 2\right) \ldots . \left(1 - \frac{1}{9} ^ 2\right) \left(1 - \frac{1}{10} ^ 2\right)$

$= \frac{3}{4} \times \frac{8}{9} \times \frac{15}{16} \times \frac{24}{25} \times \frac{35}{36} \times \frac{48}{49} \times \frac{63}{64} \times \frac{80}{81} \times \frac{99}{100}$

$= \frac{11}{20}$

The answer is $a = 11$ and $b = 20$

Jul 17, 2018

$\left(1 - \frac{1}{2} ^ 2\right) \left(1 - \frac{1}{3} ^ 2\right) \left(1 - \frac{1}{4} ^ 2\right) \ldots . \left(1 - \frac{1}{9} ^ 2\right) \left(1 - \frac{1}{10} ^ 2\right)$

$= \left(1 - \frac{1}{2}\right) \left(1 + \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \left(1 + \frac{1}{3}\right) \left(1 - \frac{1}{4}\right) \left(1 + \frac{1}{4}\right) \ldots . \left(1 - \frac{1}{9}\right) \left(1 + \frac{1}{9}\right) \left(1 - \frac{1}{10}\right) \left(1 + \frac{1}{10}\right)$

$\frac{1}{2} \cdot \frac{3}{2} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{3}{4} \cdot \frac{5}{4.} \ldots \frac{.8}{9} \cdot \frac{10}{9} \cdot \frac{9}{10} \cdot \frac{11}{10}$

$= \frac{1}{2} \cdot \frac{11}{10} = \frac{11}{20}$

So $a = 11 \mathmr{and} b = 20$