B
分析:分別表示出anan+1an+2an+3=an+an+1+an+2+an+3,an+1an+2an+3an+4=an+1+an+2+an+3+an+4,兩式相減可推斷出an+4=an,進而可知數(shù)列{an}是以4為周期的數(shù)列,只要看2010是3的多少倍,然后通過a1=-2,a2=-1,a3=1求得a4,而2010是4的502倍余2,故可知S2010=502×(-2-1+1-2)+a1+a2答案可得.
解答:依題意可知,anan+1an+2an+3=an+an+1+an+2+an+3,an+1an+2an+3an+4=an+1+an+2+an+3+an+4,
兩式相減得an+1an+2an+3(an+4-an)=an+4-an,
∵an+1an+2≠1,
∴an+4-an=0,即an+4=an,
∴數(shù)列{an}是以4為周期的數(shù)列,
∵a1a2a3a4=a1+a2+a3+a4∴a4=-2
∴S2010=502×(-2-1+1-2)+a1+a2=-2011
故選B
點評:本題主要考查了數(shù)列的遞推式和數(shù)列的求和問題.本題的關鍵是找出數(shù)列的周期性,屬于中檔題.