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(2+5+8+…+2000)-(1+4+7+…+1999)=
667
667
分析:通过观察,括号内的算式都是公差为3的等差数列,运用高斯求和公式即可求出第一个数列的项数为(2000-2)÷3+1=667,第二个数列的项数为(1999-1)÷3+1=667,代入公式求解.
解答:解:(2+2000)×[(2000-2)÷3+1]÷2-(1+1999)×[(1999-1)÷3+1]÷2,
=2002×667÷2-2000×667÷2,
=(2002-2000)×667÷2,
=2×667÷2,
=667.
故答案为:667.
点评:此题解答的关键是求出各数列的项数,然后运用公式解答即可.
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6.4÷4=7.2÷1.8=0.7÷0.35=0.72÷1.44=
0.48÷0.04=1.25÷2.5=5.5+55=16.8÷8=
0.54+2.2=3.5-0.05=2÷0.02=0.25×40=
1.5×0.6=0.32÷0.8=12.25÷0.5=73.5×0.1=
46.5+52.5=0.45×102=1.25×88=2.64+3.85+1.54=
8×1.25=5.6÷3.5=4.2÷0.7÷6=0.4×8.6×25=
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