Guess a[n]=n

When n= 1, a[n]=a[ 1]= 1.

Suppose it holds when n=k- 1(k≥2), that is, a[k- 1]=k- 1.

Then 2a [k] = 2s [k]-2s [k-1] = a [k] (a [k]+1)-a [k-1] (a [k-1]+60 [k- 1]-a[k- 1]

Answer? [k]-a[k]=a？ [k- 1]+a[k- 1]=(k- 1)？ +(k- 1)=(k- 1)(k- 1+ 1)= k(k- 1)= k？ -k

∴a？ [k]-a[k]+ 1/4=k？ -k+ 1/4, that is, (a[k]- 1/2)? =(k- 1/2)？

∴a[k]= 1/2 (k- 1/2)

That is, a[k]=k or a [k] =1-k.

∫k≥2, then 1-k < 0.

Every term of ∵{an} is positive, ∴ a [k] = k

That is, when n=k, the conjecture holds.

That is, the general formula of the sequence {an} is: a [n] = n

2. Solution:

b[n+ 1]-b[n]=3^(n+ 1)+(- 1)^n*λ*2a[n+ 1]-3^n-(- 1)^(n- 1)*λ*2a[n]

=3*3^n+(- 1)^n*λ*2(n+ 1)-3^n+(- 1)^n*λ*2n

=2*3^n+(- 1)^n*λ*(4n+2)>； 0

∴(- 1)^(n- 1)*λ<； 2*3^n/(4n+2)=3^n/(2n+ 1)

∫3n/(2n+ 1) is an increasing sequence. When the minimum value is n= 1, ∴ 3/(2+ 1)= 1, and when n=2, it is 3? /(2*2+ 1)=9/5

∴-9/5<； λ& lt； 1

And ∵ λ is a non-zero integer.

∴λ=- 1