Question

已知数列{an}的前n项和为Sn,且Sn=n-5an-85，求an和Sn

Answer 1

a1=1-5a1-85
6a1=-84
a1=-14

Sn=n-5an-85
Sn-1=(n-1)-5a(n-1)-85
an=Sn-Sn-1=n-5an-85-(n-1)+5a(n-1)+85
6an=5a(n-1)+1
6an=5a(n-1)-5+6
6(an-1)=5[a(n-1)-1]
(an-1)/[a(n-1)-1]=5/6，为定值，
a1-1=-14-1=-15
{an-1}是首项为-15，公比为5/6的等比数列。
an-1=(-15)(5/6)^(n-1)
an=(-15)(5/6)^(n-1)+1
Sn=a1+a2+...+an
=(a1-1)+(a2-1)+...+(an-1)+n
=(-15)[(1-(5/6)^n]/(1-5/6)+n
=(-90)[1-(5/6)^n]+n
=90(5/6)^n+n-90

Answer 2

1、Sn=n-5an-85
S(n+1)=n+1-5a(n+1)-85
a(n+1)=Sn(n+1)-Sn
=n+1-5a(n+1)-85-[n-5an-85 ]
=1-5a(n+1)+5an
a(n+1)=1-5a(n+1)+5an
6a(n+1)=1+5an
即a(n+1)-1=(5/6)(an-1)
又由S1=a1=1-5a1-85得a1=-14
所以{an-1}为首项-15,公比5/6的等比数列
所以an=(-15)*（5/6）^(n-1)+1

2。
Sn=(-15)*[(5/6)^0+(5/6)^1+……+(5/6)^(n-1)]+n
=[6-6*(5/6)^(n-1)]*(-15)+n