Try out this beautiful algebra problem from AMC 8, 2019 based on finding the value of the product. You may use sequential hints to solve the problem.
What is the value of the product
$\left(\frac{1 \cdot 3}{2 \cdot 2}\right)\left(\frac{2 \cdot 4}{3 \cdot 3}\right)\left(\frac{3 \cdot 5}{4 \cdot 4}\right) \cdots\left(\frac{97 \cdot 99}{98 \cdot 98}\right)\left(\frac{98 \cdot 100}{99 \cdot 99}\right) ?$
(A) $\frac{1}{2}$
(B) $\frac{50}{99}$
(C) $\frac{9800}{9801}$
(D) $\frac{100}{99}$
(E) $50$
Algebra
Value
Telescoping
Answer: is $\frac{50}{99}$
AMC 8, 2019, Problem 17
We write
$\left(\frac{1.3}{2.2}\right)\left(\frac{2.4}{3.3}\right)\left(\frac{3.5}{4.4}\right) \ldots\left(\frac{97.99}{98.98}\right)\left(\frac{98.100}{99.99}\right)$
in a different form like
$\frac{1}{2} \cdot\left(\frac{3.2}{2.3}\right) \cdot\left(\frac{4.3}{3.4}\right) \cdots \cdots \left(\frac{99.98}{98.99}\right) \cdot \frac{100}{99}$
All of the middle terms eliminate each other, and only the first and last term remains i.e.
$\frac{1}{2} \cdot \frac{100}{99}$
$\frac{1}{2} \cdot \frac{100}{99}=\frac{50}{99}$
and that is the final answer.
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