Difference between revisions of "2018 AMC 8 Problems/Problem 2"

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Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus the answer would be <math>\boxed{\textbf{(D) }7}</math>.
 
Using telescoping, most of the terms cancel out diagonally. We are left with <math>\frac{7}{1}</math> which is equivalent to <math>7</math>. Thus the answer would be <math>\boxed{\textbf{(D) }7}</math>.
  
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{{AMC8 box|year=2018|num-b=1|num-a=3}}
 
{{AMC8 box|year=2018|num-b=1|num-a=3}}
  
 
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Revision as of 17:33, 1 November 2020

Problem 2

What is the value of the product\[\left(1+\frac{1}{1}\right)\cdot\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot\left(1+\frac{1}{5}\right)\cdot\left(1+\frac{1}{6}\right)?\]

$\textbf{(A) }\frac{7}{6}\qquad\textbf{(B) }\frac{4}{3}\qquad\textbf{(C) }\frac{7}{2}\qquad\textbf{(D) }7\qquad\textbf{(E) }8$

Solution

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

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

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 $\boxed{\textbf{(D) }7}$.

See also

2018 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AJHSME/AMC 8 Problems and Solutions

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