∵a 2+b 2+c 2-ab-bc-ac
=
1
2 (2a 2+2b 2+2c 2-2ab-2bc-2ac)
=
1
2 [(a 2-2ab+b 2)+(b 2-2bc+c 2)+(c 2-2ac+a 2)]
=
1
2 [(a-b) 2+(b-c) 2+(c-a) 2]
而a=2000x+2001,b=2000x+2002,c=2000x+2003,
∴a-b=2000x+2002-(2000x+2001)=1,
同理 b-c=-1,c-a=2,
∴a 2+b 2+c 2-ab-bc-ac
=
1
2 [(a-b) 2+(b-c) 2+(c-a) 2]
=3.
故选D.