1998 AMC 8 Problem 12

Attempt Problem 12 of the 1998 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

12.

What is the value of the following expression? 2(1−12)+3(1−13)+4(1−14)+⋯+10(1−110) \begin{aligned} &2\left(1-\dfrac{1}{2}\right)+3\left(1-\dfrac{1}{3}\right) \\ &\quad {}+4\left(1-\dfrac{1}{4}\right)+\cdots \\ &\quad {}+10\left(1-\dfrac{1}{10}\right) \end{aligned}
4545
4949
5050
5454
5555
Answer: A
Concepts:summationalgebraic manipulation
Small hint:
Simplify each term n(1−1n).n(1-\frac{1}{n}).
Big hint:
The terms become consecutive integers.
Solution:
For each integer nn from 22 through 10,10,
n(1−1n)=n−1.n\left(1-\frac{1}{n}\right)=n-1.
The expression is therefore 1+2+⋯+9=45.1+2+\cdots+9=45.
Thus, the correct answer is A .
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.