Find the product of:
$(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}) \ldots(1-\frac{1}{10})$

Given :

The given expression is $(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}) \ldots(1-\frac{1}{10})$

To find :

We have to find the product of the given expression.

Solution :

$(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}) \ldots(1-\frac{1}{10})$

$1-\frac{1}{2} = \frac{2\times 1 - 1}{2} = \frac{1}{2}$

$1-\frac{1}{3} = \frac{3\times 1 - 1}{3} = \frac{2}{3}$

$1-\frac{1}{4} = \frac{4\times 1 - 1}{4} = \frac{3}{4}$

Similarly,

$1-\frac{1}{9} = \frac{9\times 1 - 1}{9} = \frac{8}{9}$

$1-\frac{1}{10} = \frac{10\times 1 - 1}{10} = \frac{9}{10}$

So, $(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}) \ldots(1-\frac{1}{10}) = (\frac{1}{2})(\frac{2}{3})(\frac{3}{4}) \ldots(\frac{8}{9})(\frac{9}{10})$

In, $ (\frac{1}{2})(\frac{2}{3})(\frac{3}{4}) \ldots(\frac{8}{9})(\frac{9}{10})$ every numbers get cancelled other than numerator of first term and denominator of last term.

Therefore,  $(\frac{1}{2})(\frac{2}{3})(\frac{3}{4}) \ldots(\frac{8}{9})(\frac{9}{10}) =\frac{1}{10}$

The product of $(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{4}) \ldots(1-\frac{1}{10})$ is $\frac{1}{10}$

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Updated on: 10-Oct-2022

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