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>1 + 2 + 3 + 4</math>      =     <math>\boxed{\textbf{(C) 10}}</math>
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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>
  
 
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 19:16, 7 February 2020

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

$\sqrt{1} + \sqrt{1+3} + \sqrt{1+3+5} + \sqrt{1+3+5+7}$ = $\sqrt{1} + \sqrt{4} + \sqrt{9} + \sqrt{16}\ = 1 + 2 + 3 + 4 = \boxed{\textbf{(C) 10}}$

Note: This comes from the fact that the sum of the first $n$ odds is $n^2$.

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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