题目内容
已知数列{an} 是通项an和公差都不为零的等差数列，设Sn=
+
+…+
，则Sn=（ ）
| 1 |
| a1a2 |
| 1 |
| a2a3 |
| 1 |
| anan+1 |
分析：由等差数列的通项公式可得an+1-an=d，可得
=
(
-
)，从而可得Sn=
+
+…+
，=
(
-
+
-
+…+
-
)
| 1 |
| anan+1 |
| 1 |
| d |
| 1 |
| an |
| 1 |
| an+1 |
| 1 |
| a1a2 |
| 1 |
| a2a3 |
| 1 |
| anan+1 |
| 1 |
| d |
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| a2 |
| 1 |
| a3 |
| 1 |
| an |
| 1 |
| an+1 |
解答：解：∵
=
(
-
)
∴Sn=
+
+…+
，
=
(
-
+
-
+…+
-
)
=
(
-
)
=
×
=
=
故选：A
| 1 |
| anan+1 |
| 1 |
| d |
| 1 |
| an |
| 1 |
| an+1 |
∴Sn=
| 1 |
| a1a2 |
| 1 |
| a2a3 |
| 1 |
| anan+1 |
=
| 1 |
| d |
| 1 |
| a1 |
| 1 |
| a2 |
| 1 |
| a2 |
| 1 |
| a3 |
| 1 |
| an |
| 1 |
| an+1 |
=
| 1 |
| d |
| 1 |
| a1 |
| 1 |
| an+1 |
=
| 1 |
| d |
| an+1-a1 |
| a1an+1 |
=
| nd |
| da1an+1 |
=
| n |
| a1an+1 |
故选：A
点评：本题主要考查了等差数列的通项公式的应用，数列求和的裂项求和方法的应用，要注意本题裂项时要乘以
是解题中容易漏掉的，这也是本题的解题关键与易错点．
| 1 |
| d |
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