Question

数列an满足a1=1,an+1=(n^2+n-入)an.(n=1,2……）,入是常数

Answer 1

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.