AMC 8 Daily Practice - Consecutive Reduction

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Problem 8 Medium

What is the value of \left(1+\frac{1}{2}\right) \times\left(1-\frac{1}{2}\right) \times\left(1+\frac{1}{3}\right) \times\left(1-\frac{1}{3}\right) \times\cdots \times\left(1+\frac{1}{10}\right) \times\left(1-\frac{1}{10}\right)?

  • A.

    \frac{11}{20}

  • B.

    \frac{1}{2}

  • C.

    \frac{3}{2}

  • D.

    \frac{9}{10}

  • E.

    \frac{21}{10}

Answer:A

We immediately notice that this problem combines two types of telescoping products.

Let's rewrite the original expression: \left(1+\frac{1}{2}\right) \times \left(1-\frac{1}{2}\right) \times \left(1+\frac{1}{3}\right) \times \left(1-\frac{1}{3}\right) \times \cdots \times \left(1+\frac{1}{10}\right) \times \left(1-\frac{1}{10}\right) = \left(1+\frac{1}{2}\right) \times \left(1+\frac{1}{3}\right) \times \cdots \times \left(1+\frac{1}{10}\right) \times \left(1-\frac{1}{2}\right) \times \left(1-\frac{1}{3}\right) \times \cdots \times \left(1-\frac{1}{10}\right)

Now we convert each term to fraction form: = \frac{3}{2} \times \frac{4}{3} \times \cdots \times \frac{11}{10} \times \frac{1}{2} \times \frac{2}{3} \times \cdots \times \frac{9}{10}

We observe two separate cancellation patterns - one in the increasing sequence and one in the decreasing sequence. For the increasing sequence: \frac{\not{3}}{2} \times \frac{\not{4}}{\not{3}} \times \cdots \times \frac{11}{\not{10}} For the decreasing sequence: \frac{1}{\not{2}} \times \frac{\not{2}}{\not{3}} \times \cdots \times \frac{\not{9}}{10}

Combining both cancellations, we are left with: = \frac{11}{2} \times \frac{1}{10}

Multiplying the remaining terms gives us the final result: = \frac{11}{20}

Final result: \boxed{\frac{11}{20}}

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