Difference between revisions of "2020 AMC 12B Problems/Problem 1"
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| − | ==Problem== | + | == Problem == |
| + | What is the value in simplest form of the following expression?<cmath>\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}</cmath> | ||
| − | |||
<math>\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}</math> | <math>\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}</math> | ||
| − | ==Solution== | + | == Solution == |
<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>\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> | ||
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== | + | == Video Solution == |
https://youtu.be/WfTty8Fe5Fo | https://youtu.be/WfTty8Fe5Fo | ||
~IceMatrix | ~IceMatrix | ||
| − | ==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}} | ||
Revision as of 00:28, 19 October 2020
Contents
Problem
What is the value in simplest form of the following expression?![\[\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}\]](http://latex.artofproblemsolving.com/3/5/9/359736c955004ea7d8e77812f4203f14ca07ba6e.png)
Solution
= 
Note: This comes from the fact that the sum of the first 
odds is 
.
Video Solution
~IceMatrix
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
