（I）因为数列{

an

λn

-(

3

λ

)n}是等差数列，公差为2所以

an+1

λn+1

-

3n+1

λn+1

=

an

λn

-

3n

λn

+2⇒an+1=λ•an+3n+1+2λn+1-λ•3n

∴b_n=3ⁿ⁺¹+2λⁿ⁺¹-λ•3ⁿ=2λⁿ⁺¹+3ⁿ（3-λ）

（II）又

lim

n→∞

bn+1

bn

=

lim

n→∞

2λn+2+3n+1(3-λ)

2λn+1+3n(3-λ)

当λ=3时，

lim

n→∞

bn+1

bn

═λ=3，

与已知矛盾，

∴λ≠3

当λ＞3时，

lim

n→∞

bn+1

bn

=

lim

n→∞

2λ+(3-λ)( 3 λ )n+1

2+ 3-λ λ ( 3 λ )n

=λ=4

∴λ=4

（III）由已知当λ=4时，

an

4n

=

3n

4n

=

11-3

4

+2(n-1)=2n⇒an=2n•4n+3n

令An=2×4+4×42+6×43++2n×4n=

8

9

+

6n-2

9

×4n+1Bn=3+32+33++3n=

3n+1

2

-

3

2

∴数列{a_n}的前n项和Sn=An+Bn=

8

9

+

6n-2

9

×4n+1+

3n+1

2

-

3

2

=-

11

18

+

3n+1

2

+

6n-2

9

×4n+1