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

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==Problem 2==
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==Problem==
What is the value of the product<cmath>\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)?</cmath>
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What is the value of the product
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<cmath>\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)?</cmath>
  
 
<math>\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</math>
 
<math>\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</math>
  
 
==Solution==
 
==Solution==
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By adding up the numbers in each of the <math>6</math> parentheses, we get:
You may simplify the expression to get <math>\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}</math>.
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<math>\frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \cdot \frac{7}{6}</math>.
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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>.
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== Video Solution (CRITICAL THINKING!!!)==
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https://youtu.be/OFsrMjvR950
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/OeaCROQV4j4
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~savannahsolver
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==Video Solution by OmegaLearn==
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https://youtu.be/TkZvMa30Juo?t=3213
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~ pi_is_3.14
  
Using telescoping, many things cancel out. You are left with <math>\frac{7}{1}</math> which is equivelant to <math>7</math>, or <math>\textbf{(D) }</math>
 
  
==See Also==
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==See also==
{{AMC8 box|year=2018|num-b=23|num-a=25}}
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{{AMC8 box|year=2018|num-b=1|num-a=3}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:19, 29 March 2023

Problem

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

Video Solution (CRITICAL THINKING!!!)

https://youtu.be/OFsrMjvR950

~Education, the Study of Everything

Video Solution

https://youtu.be/OeaCROQV4j4

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/TkZvMa30Juo?t=3213

~ pi_is_3.14


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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