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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:

Concept:

Factor each term using

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

and simplify the telescoping product.

Explanation:

Each term is

\left(1-\frac{1}{n^2}\right)=\left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)

.

So the product from

n=2

to

20

becomes:

\prod_{n=2}^{20}\left(1-\frac{1}{n^2}\right)=\prod_{n=2}^{20}\frac{n-1}{n}\cdot\prod_{n=2}^{20}\frac{n+1}{n}.

First product:

\frac{1}{2}\times\frac{2}{3}\times\cdots\times\frac{19}{20}=\frac{1}{20}.

Second product:

\frac{3}{2}\times\frac{4}{3}\times\cdots\times\frac{21}{20}=\frac{21}{2}.

Multiply them:

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

Answer:

\frac{21}{40}

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