2018 AMC 8

Complete problem set with solutions and individual problem pages

Problem 2 Easy

What is the value of the product

\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?

  • A.

    \frac 76

  • B.

    \frac 43

  • C.

    \frac 72

  • D.

    7

  • E.

    8

Answer:D

By adding up the numbers in each of the 6 parentheses, we get:

\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}.

Using telescoping, most of the terms cancel out diagonally. We are left with \frac{7}{1} which is equivalent to 7. Thus, the answer would be \boxed{\textbf{(D) }7}.

Also, you can also make the fractions \frac{7!}{6!}, which also yields \boxed{\textbf{(D)}7}

Related Problems
Problem 1 2018 AMC 8
Problem 3 2018 AMC 8
Problem 4 2018 AMC 8
Problem 5 2018 AMC 8
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