急啊``` 数列{An}中，a1=2, a (n+1)=4an-3n+1, n为N*

1.
a (n+1)=4an-3n+1
=>
a(n+1) - (n+1) = 4(an -n)
{an - n}是等比数列
2.
an-n = 4^(n-1)*(a1-1)=4^(n-1)
=>
an=4^(n-1) + n

Sn = (1+4+16+……+4^(n-1))+(1+2+3+……+n)
=(4^n - 1)/(4-1) + n(n+1)/2
=(4^n - 1)/3 ＋ n(n+1)/2

3.
S(n+1) - Sn = a(n+1) = 4^n + n + 1
S（n+1)< = 4Sn
<=>
S(n+1)-Sn <=3Sn
=>
4^n + n+1 <= 4^n - 1 + 3n(n+1)/2
<=>
2n+4 <= 3nn+3n
<=>
3nn + n - 4
=(3n+4)(n-1)>=0
原式成立