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

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== Solution ==
 
== Solution ==
We have <math></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></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>
 
Note: This comes from the fact that the sum of the first <math>n</math> odds is <math>n^2</math>.
 
Note: This comes from the fact that the sum of the first <math>n</math> odds is <math>n^2</math>.
  
== Video Solution ==
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== Video Solution (HOW TO THINK CREATIVELY!!!)==
https://youtu.be/WfTty8Fe5Fo
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https://youtu.be/kYGMJhLiU4I
 +
 
 +
~Education, the Study of Everything
 +
 
 +
 
 +
 
  
~IceMatrix
 
  
== Video Solution==
 
https://youtu.be/kYGMJhLiU4I
 
  
~Education, the Study of Everything
+
== 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 (ProblemsAnswer KeyResources)
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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