Question

已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*,求数列{an}的通项公式,求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