Therefore, when λ=2, b _ (n+1) = a _ (n+1)-2a _ n = 3 (a _ n-2a _ (n-1)) = 3b _ n.
When λ=3, b _ (n+1) = a _ (n+1)-3a _ n = 2 (a _ n-3a _ (n-1)) = 2b _ n.
So there is a real number λ that makes the sequence b_n a geometric series;
2) When λ=2, the sequence b_n is a geometric series with the first term b2=a2-2a 1=3 and the common ratio of 3.
So BN = 3 * 3 (n-2) = 3 (n- 1), that is, a _ n-2a _ (n-1) = 3 (n-1) .........................
When λ=3, the sequence b_n is a geometric series with the first term b2=a2-3a 1=2, and the common ratio is 2.
So BN = 2 * 2 (n-2) = 2 (n- 1), that is, a _ n-3a _ (n-1) = 2 (n-1) ......................... ②.
Eliminate a_(n- 1) from ① and ② to get An = 3 n-2 n.
When n= 1, a 1=3-2 also applies to the above formula.
Therefore, the general formula of sequence an is an = 3 n-2 n.
2. Because bn is arithmetic progression, let b _ n = b1+(n-1) d.
From the known n (n+1) bn = 2 (a1+...+south)
(n+ 1)(n+2)b _(n+ 1)= 2(a 1+....+(n+ 1)a_(n+ 1))
The two expressions are subtracted to get (n+1) ((n+2) b _ (n+1)-nb _ n)) = (n+1) a _ (n+1).
So a _ (n+1) = n (b _ (n+1)-bn)+2b _ (n+1) = nd+2 (b1+nd) = 2b/kloc-0.
So an is a arithmetic progression with the first term 2b 1 and the tolerance 3d.