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项和Sn

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) + nSn = (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...