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Powers and Roots Practice Test 3

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Question : 6 of 40

Marks: +1, -0

The value of

\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)

\dots\left(1-\frac{1}{19^2}\right)\left(1-\frac{1}{20^2}\right)

is equal to:

$\frac{21}{20}$
$\frac{31}{40}$
$\frac{21}{40}$
$\frac{31}{40}$

Solution:

\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\dots\left(1-\frac{1}{20^2}\right)

=\left(12-\frac{1}{2^2}\right)\left(1^2-\frac{1}{3^2}\right)\dots\left(1^2-\frac{1}{20^2}\right)

By using

(a^2-b^2)=(a-b)(a+b)

\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)

\dots\left(1-\frac{1}{20}\right)\left(1+\frac{1}{20}\right)

So,

\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\dots\left(1-\frac{1}{20}\right)

=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\dots\times\frac{19}{20}=\frac{1}{20}

...(1)

and

\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\dots\left(1+\frac{1}{20}\right)

=\frac{3}{2}\times\frac{4}{3}\times\dots\times\frac{21}{20}=\frac{21}{2}

...(2)

Now multiplying equation (1) and (2) we get,

\frac{1}{20}\times\frac{21}{2}=\frac{21}{40}

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