已知以a1為首項(xiàng)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(shí),求數(shù)列{an}的通項(xiàng)公式
(2)當(dāng)0<a1<1,c=1,d=3時(shí),試用a1表示數(shù)列{an}的前100項(xiàng)的和S100
(3)當(dāng)0<a1
1
m
(m是正整數(shù)),c=
1
m
,d≥3m時(shí),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(shí),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(shí),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(shí),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(shí),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(shí),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(shí),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(diǎn)評(píng):本題考查數(shù)列的性質(zhì)及其應(yīng)用,難度較大,解題時(shí)要認(rèn)真審題,仔細(xì)解答,避免出錯(cuò).
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