题目内容
已知数列{an}是首项为3,公差为2的等差数列,其前n项和为Sn,数列{bn}为等比数列,且b1=1,bn>0,数列{ban}是公比为64的等比数列.
(Ⅰ)求{an},{bn}的通项公式;
(Ⅱ)求证:
+
+…+
<
.
(Ⅰ)求{an},{bn}的通项公式;
(Ⅱ)求证:
1 |
S1 |
1 |
S2 |
1 |
Sn |
3 |
4 |
分析:(Ⅰ)根据等差数列的通项公式可求得an,设{bn}的公比为q,则bn=qn-1,由
=
=q2=64可求得q,从而可得bn;
(Ⅱ)由(Ⅰ)可得Sn,对
拆项后利用裂项相消法可求得
+
+…+
,从而可得结论;
ba2 |
ba1 |
b5 |
b3 |
(Ⅱ)由(Ⅰ)可得Sn,对
1 |
Sn |
1 |
S1 |
1 |
S2 |
1 |
Sn |
解答:解:(I)依题意有:an=a1+(n-1)d=3+(n-1)×2=2n+1,
设{bn}的公比为q,则bn=qn-1,
∵数列{ban}是公比为64的等比数列,
∴
=
=q2=64,解得q=8,
∴bn=8n-1;
(II)由(Ⅰ)可得,Sn=3+5+…+(2n+1)=n(n+2),
∴
=
=
(
-
),
∴
+
+…+
=
+
+
+…+
=
(1-
+
-
+
-
+…+
-
)
=
(1+
-
-
)<
.
设{bn}的公比为q,则bn=qn-1,
∵数列{ban}是公比为64的等比数列,
∴
ba2 |
ba1 |
b5 |
b3 |
∴bn=8n-1;
(II)由(Ⅰ)可得,Sn=3+5+…+(2n+1)=n(n+2),
∴
1 |
Sn |
1 |
n(n+2) |
1 |
2 |
1 |
n |
1 |
n+2 |
∴
1 |
S1 |
1 |
S2 |
1 |
Sn |
1 |
1×3 |
1 |
2×4 |
1 |
3×5 |
1 |
n(n+2) |
=
1 |
2 |
1 |
3 |
1 |
2 |
1 |
4 |
1 |
3 |
1 |
5 |
1 |
n |
1 |
n+2 |
=
1 |
2 |
1 |
2 |
1 |
n+1 |
1 |
n+2 |
3 |
4 |
点评:本题考查等差数列、等比数列的通项公式和数列求和,考查数列与不等式的综合,裂项相消法对数列求和是高考考查的重点内容,要熟练掌握.
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