Question

已知在正整数数列{a_n}中，前n项和S_n满足：S_n＝ (a_n＋2)²

(1)求证{a_n}是等差数列

(2)若b_n＝a_n－30，求数列{b_n}的前n项和的最小值．

Answer 1

答案：
解析：

(1)证明 ：由Sn＝ (an＋2)2                                                                                                                                                                                                                                                          ① 当n≥2时，Sn－1＝ (an－1＋2)2                                                                                                                                                                                                                                                ② ①－②得an＝  (an＋2)2－ (an－1＋2)2 整理得an2－an－12＝4(an＋an－1) 又an＞0 ∴an－an－1＝4． 即数列{an}构成等差数列，公差为4． (2)解 ：由Sn＝ (an＋2)2知a1＝ (a1＋2)2 即(a1－2)2＝0 ∴a1＝2 an＝a1＋(n－1)d＝4n－2 则bn＝an－30＝2n－31 令 又n∈N* ∴n＝15，此时{bn}的前n项和取得最小值． 其最小值为S15＝15b1＋·2＝－225．