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Difference between revisions of "2020 AMC 12B Problems/Problem 1"

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== Solution ==
 
== Solution ==
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<math>\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}</math> = <math>\sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}</math>
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We have <cmath>\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}.</cmath>
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Note: This comes from the fact that the sum of the first <math>n</math> odds is <math>n^2</math>.
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== Video Solution (HOW TO THINK CREATIVELY!!!)==
 +
https://youtu.be/kYGMJhLiU4I
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 +
~Education, the Study of Everything
 +
 
  
−
Note: This comes from the fact that the sum of the first <math>n</math> odds is <math>n^2</math>.
 
  
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== Video Solution ==
 
−
https://youtu.be/WfTty8Fe5Fo
 
  
−
~IceMatrix
 
  
−
== Video Solution==
 
−
https://youtu.be/kYGMJhLiU4I
 
  
−
~Education, the Study of Everything
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== Video Solution ==
 +
https://youtu.be/WfTty8Fe5Fo
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2020|ab=B|before=First Problem|num-a=2}}
 
{{AMC12 box|year=2020|ab=B|before=First Problem|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:59, 8 June 2023

Problem

What is the value in simplest form of the following expression?\[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]

$\textbf{(A) }5 \qquad \textbf{(B) }4 + \sqrt{7} + \sqrt{10} \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 4 + 3\sqrt{3} + 2\sqrt{5} + \sqrt{7}$

Solution

We have \[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7} = \sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{\textbf{(C) } 10}.\] Note: This comes from the fact that the sum of the first $n$ odds is $n^2$.

Video Solution (HOW TO THINK CREATIVELY!!!)

https://youtu.be/kYGMJhLiU4I

~Education, the Study of Everything




Video Solution

https://youtu.be/WfTty8Fe5Fo

See Also

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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 AMC 12 Problems and Solutions

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