Question

在数列{an}中,a1=2,an+1=4an-3N+1（1）证明数列{an-N}是等比数列；（2）求数列{an}的前项和

Answer 1

a[n+1]-(n+1)=4a[n]-3n+1-(n+1)=4a[n]-4n=4(a[n]-n)

所以a[n+1]-(n+1)/(a[n]-n)=4

设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