Question

已知数列{an}的前n项和为Sn，且Sn=n-5an-85，n∈N*． （1）证明：{an-1}是等

Answer 1

n=1时，
a1=S1=1-5a1-85
6a1=-84
a1=-14
n≥2时，
an=Sn-S(n-1)=n-5an-85-[(n-1)-5a(n-1)-85]
6an=5a(n-1)+1
6an-6=5a(n-1)+1-6=5a(n-1)-5
6(an -1)=5[a(n-1)-1]
(an -1)/[a(n-1)-1]=5/6，为定值
a1-1=-14-1=-15
数列{an -1}是以-15为首项，5/6为公比的等比数列。

Answer 2

Sn=n-5an-85
n=1, a1 = -12

an = Sn -S(n-1)
= 1-5an+5a(n-1)
an =(5/6)a(n-1) + 1/6
an -1 =(5/6)[a(n-1) - 1]
=>{an -1} 是等比数列, q=5/6
an -1 = (5/6)^(n-1) .(a1 -1)
= -13.(5/6)^(n-1)
an = 1 -13.(5/6)^(n-1)