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,求數(shù)列{an}的通項(xiàng)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(xiàng)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(diǎn)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(rèn)真審題,仔細(xì)解答,避免出錯.
練習(xí)冊系列答案
相關(guān)習(xí)題

科目:高中數(shù)學(xué) 來源: 題型:

已知以a1為首項(xiàng)的數(shù)列{an}滿足:an+1=
an+d,an<2
qan
 ,an≥2

(1)當(dāng)a1=1,d=1,q=
1
2
時,求數(shù)列{an}的通項(xiàng)公式;
(2)當(dāng)0<a1<1,d=1,q=
1
2
時,試用a1表示數(shù)列{an}前101項(xiàng)的和S101

查看答案和解析>>

科目:高中數(shù)學(xué) 來源: 題型:

(08年上海卷理)(18分)已知以a1為首項(xiàng)的數(shù)列{an}滿足:

⑴ 當(dāng)a1=1,c=1,d=3時,求數(shù)列{an}的通項(xiàng)公式

⑵ 當(dāng)0<a1<1,c=1,d=3時,試用a1表示數(shù)列{an}的前100項(xiàng)的和S100

⑶ 當(dāng)0<a1(m是正整數(shù)),c=d≥3m時,求證:數(shù)列a2,a3m+2,a6m+2,a9m+2成等比數(shù)列當(dāng)且僅當(dāng)d=3m

查看答案和解析>>

科目:高中數(shù)學(xué) 來源:2011-2012學(xué)年重慶市高三上學(xué)期第七次測試?yán)砜茢?shù)學(xué)試卷(解析版) 題型:解答題

已知以a1為首項(xiàng)的數(shù)列{an}滿足:an+1

⑴當(dāng)a1=1,c=1,d=3時,求數(shù)列{an}的通項(xiàng)公式

⑵當(dāng)0<a1<1,c=1,d=3時,試用a1表示數(shù)列{an}的前100項(xiàng)的和S100

⑶求證:當(dāng)0<a1(m是正整數(shù)),c=,d=3m時, a2,a3m+2,a6m+2,a9m+2成等比數(shù)列。

 

查看答案和解析>>

科目:高中數(shù)學(xué) 來源:2008年普通高等學(xué)校招生全國統(tǒng)一考試?yán)砜茢?shù)學(xué)(上海卷) 題型:解答題

(3’+7’+8’)已知以a1為首項(xiàng)的數(shù)列{an}滿足:an1=.

(1)當(dāng)a1=1,c=1,d=3時,求數(shù)列{an}的通項(xiàng)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(xiàng)的和S100;

(3)當(dāng)0<a1<(m是正整數(shù)),c=,d≥3m時,求證:數(shù)列a2-,a3m+2-,a6m+2-,a9m+2-成等比數(shù)列當(dāng)且僅當(dāng)d=3m.

 

查看答案和解析>>

同步練習(xí)冊答案