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题目 设ａ0 为常数 ,且ａｎ =3ｎ-1 - 2ａｎ-1 (ｎ ∈Ｎ+ )(Ⅰ )证明对任意ｎ ≥ 1,ａｎ =15[3ｎ+ ( - 1) ｎ-1 · 2 ｎ] + ( - 1) ｎ· 2 ｎ·ａ0 ;(Ⅱ )假设对于任意ｎ ≥ 1有ａｎ ａｎ-1 ,求ａ0 的取值范围试题是根据新教材数列一章中的一道习题设计的 ,情境新颖 ,背景公
The topic sets a0 as a constant, and an = 3n-1 - 2an-1 (n ∈ N+ )(I) proves that for any n ≥ 1, an = 15[3n+ ( - 1) n-1 · 2 n] + (- 1) n· 2 n·a0 (II) Assume that for any n ≥ 1 there is an an-1. The range of values for a0 is based on a problem in the new textbook series. The situation is novel. Background