题目内容
已知数列{an}满足an=3n-1，求证：
-
＜
+
+…+
＜
．
| n |
| 3 |
| 1 |
| 6 |
| a1 |
| a2 |
| a2 |
| a3 |
| an |
| an+1 |
| n |
| 3 |
考点：数列的求和
专题：等差数列与等比数列
分析：由已知条件得
+
+…+
=
+
+…+
＜
+
+…+
=
；
+
+…+
=
+
+…+
＞
+
+…+
，由此能证明
-
＜
+
+…+
＜
．
| a1 |
| a2 |
| a2 |
| a3 |
| an |
| an+1 |
| 2 |
| 8 |
| 8 |
| 26 |
| 3n-1 |
| 3n+1-1 |
| 3 |
| 9 |
| 9 |
| 27 |
| 3n |
| 3n+1 |
| n |
| 3 |
| a1 |
| a2 |
| a2 |
| a3 |
| an |
| an+1 |
| 2 |
| 8 |
| 8 |
| 26 |
| 3n-1 |
| 3n+1-1 |
| 2 |
| 9 |
| 8 |
| 27 |
| 3n-1 |
| 3n+1 |
| n |
| 3 |
| 1 |
| 6 |
| a1 |
| a2 |
| a2 |
| a3 |
| an |
| an+1 |
| n |
| 3 |
解答：
解：∵数列{an}满足an=3n-1，
∴
+
+…+
=
+
+…+
＜
+
+…+
=
，
+
+…+
=
+
+…+
＞
+
+…+
=
+
+…+
=（
+
+…+
）-（
+
+…+
）
=
-
=
-
(1-
)
＞
-
．
∴
-
＜
+
+…+
＜
．
∴
| a1 |
| a2 |
| a2 |
| a3 |
| an |
| an+1 |
| 2 |
| 8 |
| 8 |
| 26 |
| 3n-1 |
| 3n+1-1 |
＜
| 3 |
| 9 |
| 9 |
| 27 |
| 3n |
| 3n+1 |
| n |
| 3 |
| a1 |
| a2 |
| a2 |
| a3 |
| an |
| an+1 |
| 2 |
| 8 |
| 8 |
| 26 |
| 3n-1 |
| 3n+1-1 |
＞
| 2 |
| 9 |
| 8 |
| 27 |
| 3n-1 |
| 3n+1 |
=
| 3-1 |
| 9 |
| 9-1 |
| 27 |
| 3n-1 |
| 3n+1 |
=（
| 3 |
| 9 |
| 9 |
| 27 |
| 3n |
| 3n+1 |
| 1 |
| 32 |
| 1 |
| 33 |
| 1 |
| 3n+1 |
=
| n |
| 3 |
| ||||
1-
|
=
| n |
| 3 |
| 1 |
| 6 |
| 1 |
| 3n |
＞
| n |
| 3 |
| 1 |
| 6 |
∴
| n |
| 3 |
| 1 |
| 6 |
| a1 |
| a2 |
| a2 |
| a3 |
| an |
| an+1 |
| n |
| 3 |
点评：本题考查不等式的证明，是中档题，解题时要认真审题，注意放缩法的合理运用．
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