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Difference between revisions of "2010 AMC 10B Problems/Problem 4"

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<cmath>\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}</cmath>
 
<cmath>\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}</cmath>
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==Video Solution==
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https://youtu.be/49jID-0tszU
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-Education, the Study of Everything
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 15:59, 1 August 2022

Problem

For a real number $x$, define $\heartsuit(x)$ to be the average of $x$ and $x^2$. What is $\heartsuit(1)+\heartsuit(2)+\heartsuit(3)$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 20$

Solution

The average of two numbers, $a$ and $b$, is defined as $\frac{a+b}{2}$. Thus the average of $x$ and $x^2$ would be $\frac{x(x+1)}{2}$. With that said, we need to find the sum when we plug, $1$, $2$ and $3$ into that equation. So:


\[\frac{1(1+1)}{2} + \frac{2(2+1)}{2} + \frac{3(3+1)}{2} = \frac{2}{2} + \frac{6}{2} + \frac{12}{2} = 1+3+6= \boxed{\textbf{(C)} 10}\]

Video Solution

https://youtu.be/49jID-0tszU

-Education, the Study of Everything

Video Solution

https://youtu.be/uAc9VHtRRPg?t=209

~IceMatrix

See Also

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
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 10 Problems and Solutions

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