题目

数列1，，，…，的前n项和为　　　　　　． 　 答案：　．                                                                                             【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an====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．                                              

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