1998 AJHSME Problems/Problem 12: Difference between revisions

Revision as of 10:22, 31 July 2011

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

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

Taking the first product, we have

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

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

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

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

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

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 == 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 Problems and Solutions]]
 * [[Mathematics competition resources]]