Question

an+Sn=n.cn=an-1,求证cn是等比数列

Answer 1

a1+s1=1.得a1=1/2.

因sn=n-an.得s(n+1)=n+1-a(n+1).

a(n+1)=s(n+1)-sn=1+an-a(n+1).

a(n+1)=(1+an)/2.

c(n+1)=a(n+1)-1=(an-1)/2.

q=c(n+1)/cn=1/2.

c1=a1-1=-1/2.

所以cn是首项为-1/2,公比为1/2的等比数列