an=2a(n-1)+n-2
an+ n=2a(n-1)+2n-2
an+ n=2[a(n-1)+n-1]
(an+ n)/[a(n-1)+n-1]=2
所以数列{an+n}是等比数列
a1+1=3+1=4
an+n=(a1+1)q^(n-1)
=4*2^(n-1)
=2^(n+1)
an=2^(n+1)-n
2.
a1=2^2-1
a2=2^3-2
a3=2^4-3
.......
an=2^(n+1)-n
sn=2^2-1+2^3-2+2^4-3+................+2^(n+1)-n
=2^2+2^3+2^4+.........+2^(n+1)-1-2-3.......-n
=4(1-2^n)/(1-2)-(1+2+3+.....+n)
=4(2^n-1)-(1+n)*n/2
=2^(n+2)-n(n+1)/2-4