数列{an}满足 a1=1,a(n+1)=(n^2+n-λ)an(n=1,2 )λ是常数

(1)λ=3; a3=-3
(2)数列{an}不可能为等差数列,如（1）问题的题设,计算出a4=-27,显然,a2=-1,a3=-3,a4=-27不为等差数列
(3) 由于a1=1；a2=2-λ；a3=(6-λ)(2-λ)；a4=(12-λ)(6-λ)(2-λ);a5=(20-λ)(12-λ)(6-λ)(2-λ);.
要使得存在正整数m,当n＞m时总有an＜0,可分析出必须满足条件：a3＜0,a5＜0,a7＜0,a9＜0,.所以2＜λ＜6,12＜λ＜20,30＜λ＜42.,即n^2+n＜λ＜(n+1)^2+(n+1),其中n=1,3,5,7.