数列1，，，…，的前n项和为．

数列1，，，…，的前n项和为　　　　　　．

　．                                                                                            

【考点】数列的求和．                                                                       

【专题】等差数列与等比数列．                                                              

【分析】利用等差数列的前n项和公式化简，再利用裂项相消法求出数列的前n项和．              

【解答】解：由题意设a_n====2（），                    

所以1++…+                                                                 

=2[（1﹣）+（）+…+（）]                                            

=2（1﹣）=，                                                                       

故答案为：．                                                                              

【点评】本题考查等差数列的前n项和公式，以及裂项相消法求出数列的前n项和，注意先求数列的通项公式．