1998 AJHSME Problems/Problem 12: Difference between revisions

Latest revision as of 10:48, 20 December 2018

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

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

Taking the first product, we have

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

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

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

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

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

@@ Line 35: / Line 35: @@
 The sum of all numbers from <math>1</math> to <math>9</math> is
-<math>\frac{(9)(10)}{2}=45=\boxed{A}</math>
+<math>\frac{9\cdot10}{2}=45=\boxed{A}</math>
 == See also ==