题目内容
已知数列{an}中an=3n-2n，证明：
+
+…+
＜
（用裂项法）
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
| 3 |
| 2 |
考点：数列与不等式的综合
专题：等差数列与等比数列
分析：由已知得
=
=
•
＜
，从而
+
+…+
＜
+
•
+
•(
)2+…+
•(
)n-1=1+
+（
）2+…+（
）n-1，由此能证明
+
+…+
＜
．
| ||
|
| 3n-1-2n-1 |
| 3n-2n |
| 1 |
| 3 |
3n-
| ||
| 3n-2n |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
| 1 |
| a1 |
| 1 |
| a1 |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
| 3 |
| 2 |
解答：
证明：∵an=3n-2n，
∴
=
=1，
=
，
=
，
=
=
•
＜
，
∴
+
+…+
＜
+
•
+
•(
)2+…+
•(
)n-1
=1+
+（
）2+…+（
）n-1
=
=
(1-
)＜
．
∴
+
+…+
＜
+
•
+
•(
)2+…+
•(
)n-1
=1+
+（
）2+…+（
）n-1
=
=
(1-
)＜
．
∴
+
+…+
＜
．
∴
| 1 |
| a1 |
| 1 |
| 3-2 |
| 1 |
| an |
| 1 |
| 3n-2n |
| 1 |
| an-1 |
| 1 |
| 3n-1-2n-1 |
| ||
|
| 3n-1-2n-1 |
| 3n-2n |
=
| 1 |
| 3 |
3n-
| ||
| 3n-2n |
| 1 |
| 3 |
∴
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
| 1 |
| a1 |
| 1 |
| a1 |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| 3 |
=1+
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
=
1-
| ||
1-
|
=
| 3 |
| 2 |
| 1 |
| 3n |
| 3 |
| 2 |
∴
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
| 1 |
| a1 |
| 1 |
| a1 |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| 3 |
| 1 |
| a1 |
| 1 |
| 3 |
=1+
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 3 |
=
1-
| ||
1-
|
=
| 3 |
| 2 |
| 1 |
| 3n |
| 3 |
| 2 |
∴
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| an |
| 3 |
| 2 |
点评：本题考查不等式的证明，解题时要认真审题，注意等比数列的性质和裂项法的合理运用．
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