AMC8 1996

AMC8 1996 · Q16

AMC8 1996 · Q16. It mainly tests Arithmetic sequences basics, Casework.

$1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + 12 + 13 - \dots + 1992 + 1993 - 1994 - 1995 + 1996 = \dots$

(A) -998 -998

(B) -1 -1

(D) 1 1

(E) 998 998

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Combining in groups of four yields $1-2-3+4=0,\qquad 5-6-7+8=0,\qquad 9-10-11+12=0$ and so on. Since there are 499 groups of four in 1996, it follows that the sum is zero.

答案（C）：按每四项一组相加可得 $1-2-3+4=0,\qquad 5-6-7+8=0,\qquad 9-10-11+12=0$ 依此类推。由于 1996 中有 499 组四项，因此总和为 0。

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