1)a1=1,a2=-1
根據(jù)通項公式,a2 = (1^2+1-λ)*a1.所以,我們有-1 = (2-λ)*1,λ=3.
因此,a3 = (2^2+2-3)*a2 = 3*a2 = -3.
2)為了使得an為等差數(shù)列,我們要求d = a(n+1) - an為常數(shù).
根據(jù)通項公式,我們有,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).
為了得到等差數(shù)列,公差必須相等,所以,1-λ = (5-λ)*(2-λ),解得,λ = 3,d = -2.
將λ代入通項公式,我們有a(n+1) = (n^2+n-3)an,所以,a4 = -27.但是,a4 - a3 = -24 ≠ d.
因此,an不可能成為等差數(shù)列.