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Difference between revisions of "1998 AJHSME Problems/Problem 12"
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==Problem 12==
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==Problem==
  
 
<math>2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) + 4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)=</math>
 
<math>2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) + 4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)=</math>
Line 9: Line 9:
 
Taking the first product, we have
 
Taking the first product, we have
  
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<math>(1-\frac{1}{2})=\frac{1}{2}</math>
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<math>\left(1-\frac{1}{2}\right)=\frac{1}{2}</math>
  
 
<math>\frac{1}{2}\times2=1</math>
 
<math>\frac{1}{2}\times2=1</math>
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Looking at the second, we get
 
Looking at the second, we get
  
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<math>(1-\frac{1}{3})=\frac{2}{3}</math>
+
<math>\left(1-\frac{1}{3}\right)=\frac{2}{3}</math>
  
 
<math>\frac{2}{3}\times3=2</math>
 
<math>\frac{2}{3}\times3=2</math>
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The sum of all numbers from <math>1</math> to <math>9</math> is
 
The sum of all numbers from <math>1</math> to <math>9</math> is
  
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<math>\frac{(9)(10)}{2}=45=\boxed{A}</math>
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<math>\frac{9\cdot10}{2}=45=\boxed{A}</math>
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== See also ==
 
== See also ==
−
{{AJHSME box|year=1998|before=[[1997 AJHSME Problems|1997 AJHSME]]|after=[[1999 AMC 8 Problems|1999 AMC 8]]}}
+
{{AJHSME box|year=1998|num-b=11|num-a=13}}
 
* [[AJHSME]]
 
* [[AJHSME]]
 
* [[AJHSME Problems and Solutions]]
 
* [[AJHSME Problems and Solutions]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
 +
{{MAA Notice}}

Latest revision as of 10:48, 20 December 2018

Problem

$2\left(1-\dfrac{1}{2}\right) + 3\left(1-\dfrac{1}{3}\right) + 4\left(1-\dfrac{1}{4}\right) + \cdots + 10\left(1-\dfrac{1}{10}\right)=$

$\text{(A)}\ 45 \qquad \text{(B)}\ 49 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 54 \qquad \text{(E)}\ 55$

Solution

Taking the first product, we have

$\left(1-\frac{1}{2}\right)=\frac{1}{2}$

$\frac{1}{2}\times2=1$

Looking at the second, we get

$\left(1-\frac{1}{3}\right)=\frac{2}{3}$

$\frac{2}{3}\times3=2$

We seem to be going up by $1$.

Just to check,

$1-\frac{1}{n}=\frac{n-1}{n}$

$\frac{n-1}{n}\times n=n-1$

Now that we have discovered the pattern, we have to find the last term.

$1-\frac{1}{10}=\frac{9}{10}$

$\frac{9}{10}\times10=9$

The sum of all numbers from $1$ to $9$ is

$\frac{9\cdot10}{2}=45=\boxed{A}$

See also

1998 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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