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2008 AMC 12A Problem 4

Problem 4 of 25EasierAlgebra

Which of the following is equal to the product 84⋅128⋅1612⋯4n+44n⋯20082004? \begin{aligned} &\dfrac{8}{4} \cdot \dfrac{12}{8} \cdot \dfrac{16}{12} \\ &\quad \cdots \dfrac{4n + 4}{4n} \cdots \dfrac{2008}{2004}? \end{aligned}

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Solution

Every denominator except the first cancels with the numerator of the preceding fraction, so the product collapses to 20084=502. \dfrac{2008}{4} = 502. Thus, B is the correct answer.
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Tagged: telescoping

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