摘要：22． (1) 当n >1时, bn = Bn –Bn – 1 = –= 3n-2 令n = 1得b1＝1, ∴bn＝3n-2. 5分 (2)由an＝ (1+) a n – 1 .得 ∴an＝ 由a1 = 2 ,bn＝3n-2知. an＝(1+)(1 + )-(1+)2 ＝(1+1)(1+)-(1+) 又= = , 5分 设cn= , 当n＝1时.有(1+1) = ＞ 当n＝2时.有an＝(1+1)(1+) = = > = = cn 假设n＝k时an>cn成立.即(1+1)(1+)-(1+)>成立, 则n＝k+1时. 左边=＝ (1+1)(1+)-(1+)(1+) >(1+)= 3分 右边= c k + 1= = 由(ak+1)3 – (c k + 1)3 =– = =>0, 得ak+1 > c k + 1成立. 综合上述, an>cn对任何正整数n都成立. 3分

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