1975 AMC 12 第 6 题

Question

前八十个正偶数之和减去前八十个正奇数之和等于

The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is

0

20

40

60

80

将第

k

个偶数与第

k

个奇数配对

Pair the

k

th even number with the

k

th odd number

每一对的差

2k-(2k-1)

都相同

Each difference

2k-(2k-1)

is the same

所求差为

\begin{aligned} &(2-1)+(4-3)+\cdots\\ &\qquad +(160-159)\text{。} \end{aligned}

这是

80

个一的和，因此等于

80

。

因此，正确答案是 E。

The desired difference is

\begin{aligned} &(2-1)+(4-3)+\cdots\\ &\qquad +(160-159). \end{aligned}

It is the sum of

80

ones, hence

80.

Therefore, the correct answer is E.

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