Difference between revisions of "2020 AMC 12B Problems/Problem 1"
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== 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>\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>. | ||
Revision as of 12:15, 22 August 2022
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
Video Solution
~Education, the Study of Everything
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. 
