(Ⅰ)由已知,得(a n+1-a n)•8(a n-1)+(a n-1) 2=0.
即(a n-1)(8a n+1-7a n-1)=0.
∵a 1=2≠1,∴a 2≠1,同理a 3≠1,…,a n≠1.
∴8a n+1=7a n+1.
即8(a n+1-1)=7(a n-1),
∴数列{a n-1}是以a 1-1=1为首项,
7
8 为公比的等比数列.
(Ⅱ)由(1),得 a n -1=(
7
8 ) n-1 .
∴ b n =(n+1)•(
7
8 ) n .
则 b n+1 =(n+2)•(
7
8 ) n+1 .
∵
b n+1
b n =
n+2
n+1 •
7
8 ,设
b n+1
b n ≥1,则n≤6.
因此,当n<6时,b n<b n+1;当n=6时,b 6=b 7,当n>6时,b n>b n+1.
∴当n=6或7时,b n取得最大值.
(Ⅲ) S n =2•
7
8 +3•(
7
8 ) 2 +4•(
7
8 ) 3 +…+n•(
7
8 ) n-1 +(n+1)•(
7
8 ) n
7
8 • S n =2•(
7
8 ) 2 +3•(
7
8 ) 3 +4•(
7
8 ) 4 +…+n•(
7
8 ) n +(n+1)•(
7
8 ) n+1
相减得:
1
8 • S n =2•
7
8 +(
7
8 ) 2 +(
7
8 ) 3 +…+(
7
8 ) n -(n+1)•(
7
8 ) n+1 =
7
8 +
7
8 ×8×[1-(
7
8 ) n ]-(n+1)•(
7
8 ) n+1
=
63
8 -(n+9)•(
7
8 ) n+1
∴ S n =63-8(n+9)•(
7
8 ) n+1 .