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Powers and Roots Practice Test 2

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Question : 28 of 40

Marks: +1, -0

The value of

\left(1^2-\frac{1}{3^2}\right)\left(1^2-\frac{1}{4^2}\right)\left(1^2-\frac{1}{5^2}\right)

\ldots\left(1^2-\frac{1}{100^2}\right)

is equal to:

$\frac{101}{150}$
$\frac{7}{150}$
$\frac{9}{200}$
$\frac{3}{100}$

Solution:

\left(1^2-\frac{1}{3^2}\right)\left(1^2-\frac{1}{4^2}\right)\left(1^2-\frac{1}{5^2}\right)

\ldots\left(1^2-\frac{1}{100^2}\right)

By using

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

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

\left(1-\frac{1}{4}\right)\left(1+\frac{1}{4}\right)\left(1-\frac{1}{5}\right)

\left(1+\frac{1}{5}\right)\ldots\left(1-\frac{1}{100}\right)\left(1+\frac{1}{100}\right)

Now

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

\left(1-\frac{1}{5}\right)\ldots\left(1-\frac{1}{100}\right)

=\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times\ldots\times\frac{9}{100}

=\frac{2}{100}

......(i)

\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)\left(1+\frac{1}{5}\right)\ldots\left(1+\frac{1}{100}\right)

=\frac{4}{3}\times\frac{5}{4}\times\frac{6}{5}\times\ldots\times\frac{101}{100}=\frac{101}{3}

......(ii)

Now multiplying equation (i) and (ii) we get,

=\frac{2}{100}\times\frac{101}{3}=\frac{101}{150}

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