题目内容
数列{an}的通项公式an=n（n+1），Sn为数列{
}的前n项和，则Sn=
．
| 1 |
| an |
| n |
| n+1 |
| n |
| n+1 |
分析：由已知可得：
=
=
-
，利用“裂项求和”即可得出．
| 1 |
| an |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
解答：解：∵
=
=
-
，
∴Sn=(1-
)+(
-
)+…+(
-
)
=1-
=
．
故答案为
．
| 1 |
| an |
| 1 |
| n(n+1) |
| 1 |
| n |
| 1 |
| n+1 |
∴Sn=(1-
| 1 |
| 2 |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| n |
| 1 |
| n+1 |
=1-
| 1 |
| n+1 |
=
| n |
| n+1 |
故答案为
| n |
| n+1 |
点评：熟练掌握“裂项求和”是解题的关键．
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