已知数列an中,a1=2,an+1=4an-3n+1,bn=an-n,求证数列bn为等比数列,求an前n项和

an+1=4an-3n+1
an+1-n-1=4an-4n
an+1-（n+1)=4(an-n)
[an+1-(n+1)]/(an-n)=4
所以an-n是等比数列
bn=an-n,所以bn是等比数列,b1=1,q=4
Sbn=(4^n-1)/3
San=Sbn-(1+2+3.+n)
=(4^n-1)/3-(1+n)*n/2
=(4^n-1)/3-(n+n^2)/2