在数列{an}中,a1=2,a(n+1)=4an-3n+1.（1）证明{an-n}是等比数列 (2)求数列{an}的前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)
a(n+1)=4^n + n+1>0
所以S(n+1)>Sn