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        1. 已知以a1為首項的數(shù)列{an}滿足:an+1=
          an+can<3
          an
          d
          an≥3

          (1)當(dāng)a1=1,c=1,d=3時,求數(shù)列{an}的通項公式
          (2)當(dāng)0<a1<1,c=1,d=3時,試用a1表示數(shù)列{an}的前100項的和S100
          (3)當(dāng)0<a1
          1
          m
          (m是正整數(shù)),c=
          1
          m
          ,d≥3m時,求證:數(shù)列a2-
          1
          m
          ,a3m+2-
          1
          m
          ,a6m+2-
          1
          m
          ,a9m+2-
          1
          m
          成等比數(shù)列當(dāng)且僅當(dāng)d=3m.
          分析:(1)由題意得an=
          1,n=3k-2
          2,n=3k-1
          3,n=3k
          ,(k∈Z+)

          (2)由題意知a3k-1=
          a1
          33k-1
          +1
          ,a3k=
          a1
          33k-1
          +2
          ,a3k+1=
          a1
          33k-1
          +3
          ,所以S100=a1+(a2+a3+a4)+(a5+a6+a6)+…+(a98+a99+a100)=a1+(3a1+6)+(a1+6)+(
          a1
          3
          +6)+…+(
          a1
          331
          +6)
          =
          1
          2
          (11-
          1
          331
          )a1+198

          (3)由題設(shè)條件可知,當(dāng)d=3m時,數(shù)列a2-
          1
          m
          ,a3m+2-
          1
          m
          ,a6m+2-
          1
          m
          ,a9m+2-
          1
          m

          是公比為
          1
          3m
          的等比數(shù)列;當(dāng)d≥3m+1時,a3m+2-
          1
          m
          <0
          ,a6m+2-
          1
          m
          >0
          ,a9m+2-
          1
          m
          >0

          故數(shù)列a2-
          1
          m
          a3m+2-
          1
          m
          ,a6m+2-
          1
          m
          a9m+2-
          1
          m
          ,不是等比數(shù)列.所以,數(shù)列a2-
          1
          m
          ,a3m+2-
          1
          m
          ,a6m+2-
          1
          m
          a9m+2-
          1
          m
          ,成等比數(shù)列當(dāng)且僅當(dāng)d=3m
          解答:解:(1)由題意得an=
          1,n=3k-2
          2,n=3k-1
          3,n=3k
          ,(k∈Z+)

          (2)當(dāng)0<a1<1時,a2=a1+1,a3=a1+2,a4=a1+3,
          a5=
          a1
          3
          +1
          a6=
          a1
          3
          +2
          ,a7=
          a1
          3
          +3
          ,a3k-1=
          a1
          33k-1
          +1
          a3k=
          a1
          33k-1
          +2
          ,a3k+1=
          a1
          33k-1
          +3

          ∴S100=a1+(a2+a3+a4)+(a5+a6+a7)+…+(a98+a99+a100
          =a1+(3a1+6)+(a1+6)+(
          a1
          3
          +6)+…+(
          a1
          331
          +6)

          =a1+a1(3+1+
          1
          3
          +…+
          1
          331
          )+6×33

          =
          1
          2
          (11-
          1
          331
          )a1+198

          (3)當(dāng)d=3m時,a2=a1+
          1
          m
          ,
          a3m=a1+
          3m-1
          m
          =a1-
          1
          m
          +3<3<a1+3=a 3m+1
          ,
          a3m+2=
          a1
          3m
          +
          1
          m
          ;
          a6m=
          a1
          3m
          -
          1
          m
          +3<3<
          a1
          3m
          +3=a6m+1

          a6m+2=
          a1
          9m2
          +
          1
          m
          ;
          a9m=
          a1
          9m2
          -
          1
          m
          +3<3<
          a1
          9m2
          +3=a9m+1
          ,
          a9m+2=
          a1
          27m3
          +
          1
          m

          a2-
          1
          m
          =a1
          ,a3m+2-
          1
          m
          =
          a1
          3m
          ,a6m+2-
          1
          m
          =
          a1
          9m2

          a9m+2-
          1
          m
          =
          a1
          27m3

          綜上所述,當(dāng)d=3m時,數(shù)列a2-
          1
          m
          ,a3m+2-
          1
          m
          a6m+2-
          1
          m
          ,a9m+2-
          1
          m

          是公比為
          1
          3m
          的等比數(shù)列
          當(dāng)d≥3m+1時,a3m+2=
          a1+3
          d
          ∈(0,
          1
          m
          )
          ,
          a6m+2=
          a1+3
          d
          +3∈(3,3+
          1
          m
          )
          ,
          a6m+3=
          a1+3
          d
          +3
          d
          ∈(0,
          1
          m
          )
          ,
          a9m+2=
          a1+3
          d
          +3
          d
          +
          3m-1
          m
          ∈(3-
          1
          m
          ,3)
          ,
          由于a3m+2-
          1
          m
          <0
          ,a6m+2-
          1
          m
          >0
          a9m+2-
          1
          m
          >0

          故數(shù)列a2-
          1
          m
          ,a3m+2-
          1
          m
          a6m+2-
          1
          m
          ,a9m+2-
          1
          m
          ,不是等比數(shù)列
          所以,數(shù)列a2-
          1
          m
          ,a3m+2-
          1
          m
          a6m+2-
          1
          m
          ,a9m+2-
          1
          m

          成等比數(shù)列當(dāng)且僅當(dāng)d=3m
          點評:本題考查數(shù)列的性質(zhì)及其應(yīng)用,難度較大,解題時要認真審題,仔細解答,避免出錯.
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