The value of left(2-frac{1}{3}right)left(2-fr | vedclass

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Question

(A) The given expression is $\left(2-\frac{1}{3}\right)\left(2-\frac{3}{5}\right)\left(2-\frac{5}{7}\right) \ldots \left(2-\frac{999}{1001}\right)$.

Vedclass · Accepted Answer

$\frac{1003}{3}$

Vedclass · Answer

(A) The given expression is $\left(2-\frac{1}{3}ight)\left(2-\frac{3}{5}ight)\left(2-\frac{5}{7}ight) \ldots \left(2-\frac{999}{1001}ight)$.
Simplifying each term:
$\left(\frac{6-1}{3}ight)\left(\frac{10-3}{5}ight)\left(\frac{14-5}{7}ight) \ldots \left(\frac{2002-999}{1001}ight)$
$= \frac{5}{3} 	imes \frac{7}{5} 	imes \frac{9}{7} 	imes \ldots 	imes \frac{1003}{1001}$.
Observing the pattern,the numerator of each fraction cancels out the denominator of the subsequent fraction:
$= \frac{{5}}{3} 	imes \frac{{7}}{{5}} 	imes \frac{{9}}{{7}} 	imes \ldots 	imes \frac{1003}{{1001}}$.
After cancellation,we are left with the numerator of the last term and the denominator of the first term:
$= \frac{1003}{3}$.

The value of $\left(2-\frac{1}{3}\right)\left(2-\frac{3}{5}\right)\left(2-\frac{5}{7}\right) \ldots \left(2-\frac{999}{1001}\right)$ is

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