Question

已知数列｛an｝的前n项和为Sn，且Sn=n-5an-85，n属于N*，证明：｛an-1｝是等比数列

Answer 1

a(1)=s(1)=1-5a(1)-85,6a(1)=-84,a(1)=-14. a(n+1)=s(n+1)-s(n)=(n+1)-5a(n+1)-85-[n-5a(n)-85]=1-5a(n+1)+5a(n), 6a(n+1)=5a(n)+1,6a(n+1)-6=5a(n)-5, 6[a(n+1)-1]=5[a(n)-1], a(n+1)-1=(5/6)[a(n)-1], {a(n)-1}是首项为a(1)-1=-14-1=-15,公比为(5/6)的等比数列. a(n)-1=(-15)(5/6)^(n-1),n=1,2,... a(n)=1-15(5/6)^(n-1),n=1,2,...