数列an满足a1=1,an+1=(n^2+n-入)an.(n=1,2……),入是常数1)当a2=-1时,求入及a3的值2)数列an是否可能为等差数列?若可能,求出它的通项公式;若不可能,说明理由.
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![数列an满足a1=1,an+1=(n^2+n-入)an.(n=1,2……),入是常数1)当a2=-1时,求入及a3的值2)数列an是否可能为等差数列?若可能,求出它的通项公式;若不可能,说明理由.](/uploads/image/z/4050697-49-7.jpg?t=%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3a1%3D1%2Can%2B1%3D%28n%5E2%2Bn-%E5%85%A5%29an.%28n%3D1%2C2%E2%80%A6%E2%80%A6%EF%BC%89%2C%E5%85%A5%E6%98%AF%E5%B8%B8%E6%95%B01%29%E5%BD%93a2%3D-1%E6%97%B6%2C%E6%B1%82%E5%85%A5%E5%8F%8Aa3%E7%9A%84%E5%80%BC2%29%E6%95%B0%E5%88%97an%E6%98%AF%E5%90%A6%E5%8F%AF%E8%83%BD%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%3F%E8%8B%A5%E5%8F%AF%E8%83%BD%2C%E6%B1%82%E5%87%BA%E5%AE%83%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%E8%8B%A5%E4%B8%8D%E5%8F%AF%E8%83%BD%2C%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1.)
数列an满足a1=1,an+1=(n^2+n-入)an.(n=1,2……),入是常数1)当a2=-1时,求入及a3的值2)数列an是否可能为等差数列?若可能,求出它的通项公式;若不可能,说明理由.
数列an满足a1=1,an+1=(n^2+n-入)an.(n=1,2……),入是常数
1)当a2=-1时,求入及a3的值
2)数列an是否可能为等差数列?若可能,求出它的通项公式;若不可能,说明理由.
数列an满足a1=1,an+1=(n^2+n-入)an.(n=1,2……),入是常数1)当a2=-1时,求入及a3的值2)数列an是否可能为等差数列?若可能,求出它的通项公式;若不可能,说明理由.
1)a1=1,a2=-1
根据通项公式,a2 = (1^2+1-λ)*a1.所以,我们有-1 = (2-λ)*1,λ=3.
因此,a3 = (2^2+2-3)*a2 = 3*a2 = -3.
2)为了使得an为等差数列,我们要求d = a(n+1) - an为常数.
根据通项公式,我们有,a(n+1) - an = (n^2+n-λ-1)an.
已知a1 = 1,所以,a2 = 2-λ,d = a2 - a1 = 1-λ.
a3 = (2^2+2-λ)a2 = (6-λ)(2-λ),d = a3 - a2 = (5-λ)*(2-λ).
为了得到等差数列,公差必须相等,所以,1-λ = (5-λ)*(2-λ),解得,λ = 3,d = -2.
将λ代入通项公式,我们有a(n+1) = (n^2+n-3)an,所以,a4 = -27.但是,a4 - a3 = -24 ≠ d.
因此,an不可能成为等差数列.
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