AMC8 1998
AMC8 1998 · Q12
AMC8 1998 · Q12. It mainly tests Fractions.
$2(1 - \frac{1}{2}) + 3(1 - \frac{1}{3}) + 4(1 - \frac{1}{4}) + \dots + 10(1 - \frac{1}{10}) =$
$2(1 - \frac{1}{2}) + 3(1 - \frac{1}{3}) + 4(1 - \frac{1}{4}) + \dots + 10(1 - \frac{1}{10}) =$
(A)
45
45
(B)
49
49
(C)
50
50
(D)
54
54
(E)
55
55
Answer
Correct choice: (A)
正确答案：（A）
Solution
Answer (A):
$2\left(1-\frac{1}{2}\right)+3\left(1-\frac{1}{3}\right)+4\left(1-\frac{1}{4}\right)+\cdots+10\left(1-\frac{1}{10}\right)=$
$2\left(\frac{1}{2}\right)+3\left(\frac{2}{3}\right)+4\left(\frac{3}{4}\right)+\cdots+10\left(\frac{9}{10}\right)=$
$1+2+3+\cdots+9=45.$
答案（A）：
$2\left(1-\frac{1}{2}\right)+3\left(1-\frac{1}{3}\right)+4\left(1-\frac{1}{4}\right)+\cdots+10\left(1-\frac{1}{10}\right)=$
$2\left(\frac{1}{2}\right)+3\left(\frac{2}{3}\right)+4\left(\frac{3}{4}\right)+\cdots+10\left(\frac{9}{10}\right)=$
$1+2+3+\cdots+9=45.$
Topics
Related Questions
Practice full AMC exams on amcdrill.
Try full-length practice and diagnostics at www.amcdrill.com.