已知数列{an}满足an+Sn=n,数列{bn}满足b1=a1,且bn=an-a(n-1),(n≥2）,试求数列{bn}的前n项的和Tn
问题描述：
已知数列{an}满足an+Sn=n,数列{bn}满足b1=a1,且bn=an-a(n-1),(n≥2）,试求数列{bn}的前n项的和Tn
答
an+Sn=n
a(n-1)+S(n-1)=n-1
an-a(n-1)+Sn-S(n-1)=1
an-a(n-1)+an=1
bn+an=1
b1+a1+b2+a2+...+bn+an=n
Tn+Sn=n
Tn=n-Sn=n-(n-an)=an
∴Tn=an
答
an+Sn=n,a(n-1)+S(n-1)=n-1,前式减后式得：an-a(n-1)+an=1,2an-a(n-1)=1；2(an-1)=a(n-1)-1,(an-1)/[a(n-1)-1]=1/2,则数列{an-1}为等比数列公比q=1/2,因为a1+a1=1,首项为a1-1=-1/2,通项：an-1=-1/2*1/2^(n-1)=-1/2^n...