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题目已知数列{a_n}满足:a_1=2,a_n=2(a_(n-1)+n)(n=2,3,…).求数列{a_n}的通项公式.(2013年全国高中数学联赛(B卷)试题)本文从一题多解,一题多变两个角度对本题目进行探究,希望对同仁有所帮助.一、一题多解解法1:a_1=2,a_2=2(a_1+2)=8,当n≥3时,我们有a_n-2a_(n-1)=2n,a_n-1-2a_(n-2)=2(n-1),两式相减,得a_n-3a_(n-1)+2a_(n-2)=2,即a_n-a_(n-1)+2=2(a_(n-1)-a_(n-2)+2),令b_n=a_n-a_(n-1)+2(n≥2),则数列{b_n}(n≥2)是公比为2的等比数列,且b_2=
The topic known sequence {a_n} satisfies: a_1=2, a_n=2(a_(n-1)+n)(n=2,3,...). Find the general term formula for the sequence {a_n}. (National High School Mathematics League (Volume B) Questions) This article explores this topic from two perspectives, one subject with multiple variations, and one that hopes to help colleagues. First, a multi-solution solution 1:a_1=2,a_2= 2(a_1+2)=8, when n≥3, we have a_n-2a_(n-1)=2n, a_n-1-2a_(n-2)=2(n-1), and the two equations are subtracted. , get a_n-3a_(n-1)+2a_(n-2)=2, namely a_n-a_(n-1)+2=2(a_(n-1)-a_(n-2)+2) Let b_n=a_n-a_(n-1)+2(n≥2), then the sequence {b_n}(n≥2) is a geometric sequence with a common ratio of 2, and b_2=