题目内容
求和:
+
+
+…+
.
| 1 |
| 1×5 |
| 1 |
| 5×9 |
| 1 |
| 9×13 |
| 1 |
| (4n-3)(4n+1) |
考点:数列的求和
专题:等差数列与等比数列
分析:由
=
(
-
),利用“裂项求和”即可得出.
| 1 |
| (4n-3)(4n+1) |
| 1 |
| 4 |
| 1 |
| 4n-3 |
| 1 |
| 4n+1 |
解答:
解:∵
=
(
-
).
∴原式=
[(1-
)+(
-
)+…+(
-
)]
=
(1-
)
=
.
| 1 |
| (4n-3)(4n+1) |
| 1 |
| 4 |
| 1 |
| 4n-3 |
| 1 |
| 4n+1 |
∴原式=
| 1 |
| 4 |
| 1 |
| 5 |
| 1 |
| 5 |
| 1 |
| 9 |
| 1 |
| 4n-3 |
| 1 |
| 4n+1 |
=
| 1 |
| 4 |
| 1 |
| 4n+1 |
=
| n |
| 4n+1 |
点评:本题考查了“裂项求和”方法,属于基础题.
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