a3 = 2 a2-3+2 = 2 a2- 1 = 2×4- 1 = 7

When n≥2,

an=2a(n- 1)-n+2

an-n = 2a(n- 1)-2n+2 = 2a(n- 1)-2(n- 1)= 2[a(n- 1)-(n- 1)]

(an-n)/[a (n-1)-(n-1)] = 2, which is a fixed value.

A 1-1= 2-1=1,and the sequence {an-n} is a geometric series with1as the first term and 2 as the common ratio.

an-n= 1×2^(n- 1)=2^(n- 1)

an=n+2^(n- 1)

bn=an/2^(n- 1)=[n+2^(n- 1)]/2^(n- 1)= 1+ n/2^(n- 1)

Sn=b 1+b2+...+bn = 1+ 1/ 1+ 1+2/2+...+ 1+n/2^(n- 1)=n+ 1/ 1+2/2...+n/2^(n- 1)

Let cn =11+2/2+...+n/2 (n-1)

Then (1/2) cn =1/2+2+...+(n-1)/2 (n-1)+n/2?

Cn-( 1/2)Cn =( 1/2)Cn = 1+ 1/2+...+ 1/2^(n- 1)-n/2？

= 1×[ 1-( 1/2)? ]/( 1- 1/2)-n/2？

=2- (n+2)/2？

Cn=4-2(n+2)/2？ = 4-n/2^(n- 1)- 1/2^(n-2)

sn = n+cn = n+4-n/2^(n- 1)- 1/2^(n-2)