• 等差数列{an}的前n项和为Sn,已知(a1006-1)3+2013(a1006-1)=1,(a1008-1)3+2013(a1008-1)=-1,则( )试题及答案-单选题-云返教育

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      等差数列{an}的前n项和为Sn,已知(a1006-1)3+2013(a1006-1)=1,(a1008-1)3+2013(a1008-1)=-1,则(  )

      试题解答


      B
      解:∵(a1006-1)3+2013(a1006-1)=1>0,(a1008-1)3+2013(a1008-1)=-1<0,
      ∴a
      1006>1,a1008<1,即a1008<a1006
      设a=a
      1006-1,b=a1008-1,
      则a>0,b<0,
      则条件等价为:a
      3+2013a=1,b3+2013b=-1,
      两式相加得a
      3+b3+2013(a+b)=0,
      即(a+b)(a
      2-ab+b2)+2013(a+b)=0,
      ∴(a+b)(a
      2-ab+b2+2013)=0,
      ∵a>0,b<0,
      ∴ab<0,-ab>0,
      即a
      2-ab+b2+2013>0,
      ∴必有a+b=0,
      即a
      1006-1+a1008-1=0,
      ∴a
      1006+a1008=2,即a1006+a1008=a1+a2013=2,
      ∴S
      2013=
      2013(a1+a2013)
      2
      =2013,
      故选:B.
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