Question

已知数列{an}的前n项和为Sn,且对任意n属于N+有an+Sn=n,设Cn=n(1-bn)求数列{Cn}的前n项和Tn

Answer 1

(1)an+Sn=n

a(n+1)+S(n+1)=n+1两式相减2a(n+1)-an=1,即2(a(n+1)-1)=an-1,2b(n+1)=bn

而a1+a1=1,a1=1/2,b1=-1/2,{bn}是以-1/2为首项,1/2为公比的等比数列

(2)bn=-1/2^n,Cn=n(1-bn)=n+n/2^n

而n/2^n=(n+1)/2^(n-1)-(n+2)/2^n利用错位相减得

Tn=n(n+1)/2+2-(n+2)/2^n