AMC12 2003 A

AMC12 2003 A · Q1

AMC12 2003 A · Q1. It mainly tests Arithmetic sequences basics, Arithmetic misc.

What is the difference between the sum of the first $2003$ even counting numbers and the sum of the first $2003$ odd counting numbers?

前 $2003$ 个偶计数数之和与前 $2003$ 个奇计数数之和的差是多少？

(A) 0 0

(B) 1 1

(D) 2003 2003

(E) 4006 4006

Answer

Correct choice: (D)

正确答案：（D）

Solution

The first $2003$ even counting numbers are $2,4,6,...,4006$. The first $2003$ odd counting numbers are $1,3,5,...,4005$. Thus, the problem is asking for the value of $(2+4+6+...+4006)-(1+3+5+...+4005)$. $(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$ $= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}$ The answer is 2003

前 $2003$ 个偶计数数是 $2,4,6,...,4006$。前 $2003$ 个奇计数数是 $1,3,5,...,4005$。因此，本题要求的是 $(2+4+6+...+4006)-(1+3+5+...+4005)$ 的值。 $(2+4+6+...+4006)-(1+3+5+...+4005) = (2-1)+(4-3)+(6-5)+...+(4006-4005)$ $= 1+1+1+...+1 = \boxed{\mathrm{(D)}\ 2003}$ 答案是 2003

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