Question

数列{an}的前n项和为Sn,已知Sn+an=-n（n∈N）恒成立.①求数列an的通项公式.②bn=ln（an+1）,求

Answer 1

Sn+an=-n

a1+a1=-1,a1=-1/2

n>=2:

S(n-1)+a(n-1)=-(n-1)

二式相减得到：an+an-a(n-1)=-1

2an=a(n-1)-1

2(an+1)=a(n-1)+1

即数列{an+1}是以a1+1=1/2为首项,1/2为公比的等比数列,则有an+1=(1/2)^n

即有an=(1/2)^n-1

(2)bn=ln(an+1)=ln(1/2)^n=nln1/2

cn=anbn=[(1/2)^n-1]nln1/2=ln1/2*n*(1/2)^n-nln1/2

Tn=ln1/2[1*1/2+2*(1/2)^2+3*(1/2)^3+...+n*(1/2)^n]-ln1/2*(1+2+...+n)

设An=1*1/2+2*(1/2)^2+3*(1/2)^3+,.+n*(1/2)^n

1/2An=1*(1/2)^2+2*(1/2)^3+...+n*(1/2)^(n+1)

1/2An=An-1/2An=1/2+(1/2)^2+(1/2)^3+...+(1/2)^n-n*(1/2)^(n+1)=1/2*(1-1/2^n)/(1-1/2)-n*1/2^n*1/2

故An=2-2*1/2^n-n*1/2^n

故有Tn=ln1/2[2-(2+n)*1/2^n]-ln1/2*(1+n)n/2