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

## 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

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