Test Index

Powers and Roots Practice Test 1

© examsnet.com

Question : 18 of 40

Marks: +1, -0

The product 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}{11^2}\right)\left(1-\frac{1}{12^2}\right)

is equal to:

$\frac{5}{8}$
$\frac{11}{20}$
$\frac{17}{18}$
$\frac{13}{24}$

Solution: 👈: Video Solution

\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}{11^2}\right)\left(1-\frac{1}{12^2}\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)\left(1-\frac{1}{4}\right)

\left(1+\frac{1}{4}\right)\dots\left(1-\frac{1}{12}\right)\left(1+\frac{1}{12}\right)

Here

\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\dots\left(1-\frac{1}{12}\right)

First

=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times

\dots\times\frac{11}{12}=\frac{1}{12}

...(i)

Now

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

\dots\left(1+\frac{1}{12}\right)

=\frac{2}{3}\times\frac{4}{3}\times\frac{5}{4}\times\dots\times\frac{13}{12}

=\frac{13}{2}

......(ii)

Multiplying equation (i) and (ii), we get

\frac{1}{12}\times\frac{13}{2}=\frac{13}{24}

© examsnet.com

Go to Question:

Prev Question

Next Question