题目内容
已知an=n+2，设bn=
，Sn为数列{bn}的前n项和，求证：
≤Sn≤
．
| 2an+1 |
| an(an+1)(an+2) |
| 7 |
| 60 |
| 13 |
| 24 |
考点：数列与不等式的综合,数列的求和
专题：综合题,等差数列与等比数列
分析：bn=
=
+
=
（
-
）+（
-
），求和，即可证明结论．
| 2an+1 |
| an(an+1)(an+2) |
| 1 |
| (n+2)(n+4) |
| 1 |
| (n+3)(n+4) |
| 1 |
| 2 |
| 1 |
| n+2 |
| 1 |
| n+4 |
| 1 |
| n+3 |
| 1 |
| n+4 |
解答：
证明：an=n+2，bn=
=
+
=
（
-
）+（
-
）
∴Sn=
（
-
+
-
+…+
-
）+（
-
+
-
+…+
-
）
=
（
+
-
-
）+（
-
）=
-
-
，
∴n=1时，Sn=
，
∴
≤Sn≤
．
| 2an+1 |
| an(an+1)(an+2) |
| 1 |
| (n+2)(n+4) |
| 1 |
| (n+3)(n+4) |
| 1 |
| 2 |
| 1 |
| n+2 |
| 1 |
| n+4 |
| 1 |
| n+3 |
| 1 |
| n+4 |
∴Sn=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| 4 |
| 1 |
| 6 |
| 1 |
| n+2 |
| 1 |
| n+4 |
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 6 |
| 1 |
| n+3 |
| 1 |
| n+4 |
=
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n+3 |
| 1 |
| n+4 |
| 1 |
| 4 |
| 1 |
| n+4 |
| 13 |
| 24 |
| 1 |
| n+3 |
| 2 |
| n+4 |
∴n=1时，Sn=
| 7 |
| 60 |
∴
| 7 |
| 60 |
| 13 |
| 24 |
点评：本题考查数列的求和，考查数列与不等式的综合，正确求和是关键．
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