Difference between revisions of "1998 AJHSME Problems/Problem 12"

(Solution)
Line 9: Line 9:
 
Taking the first product, we have
 
Taking the first product, we have
  
<math>(1-\frac{1}{2})=\frac{1}{2}</math>
+
<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>
Line 36: Line 36:
  
 
<math>\frac{(9)(10)}{2}=45=\boxed{A}</math>
 
<math>\frac{(9)(10)}{2}=45=\boxed{A}</math>
 
  
 
== See also ==
 
== See also ==

Revision as of 10:47, 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

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

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

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

$\frac{(9)(10)}{2}=45=\boxed{A}$

See also

1998 AJHSME (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS