题目内容
已知数列an的通项公式为an=| n+1 |
| 2 |
| 1 |
| a1•a3 |
| 1 |
| a2•a4 |
| 1 |
| an•an+2 |
分析：由于
=
= 2(
-
)，适合用裂项求和．
| 1 |
| an•an+2 |
| 4 |
| (n+1)(n+3) |
| 1 |
| n+1 |
| 1 |
| n+3 |
解答：解：∵an=
∴
=
=2(
-
)
∴Tn=2(
-
+
-
+…+
-
)
=2(
+
-
-
)=
| n+1 |
| 2 |
∴
| 1 |
| anan+2 |
| 4 |
| (n+1)(n+3) |
| 1 |
| n+1 |
| 1 |
| n+3 |
∴Tn=2(
| 1 |
| 2 |
| 1 |
| 4 |
| 1 |
| 3 |
| 1 |
| 5 |
| 1 |
| n+1 |
| 1 |
| n+3 |
=2(
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n+2 |
| 1 |
| n+3 |
| 5n2+25n+24 |
| 3(n+2)(n+3) |
点评：本题主要考查裂项求和，解题时要注意裂项相消后，判断剩余几项．
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