已知Sn为数列{an}的前项n和,且Sn=2an+n^2-3n-2(n∈N*),令bn=an-2n(n∈N*)

(1) Sn=2an+n^2-3n-2 S(n-1)=2a(n-1)+(n-1)^2-3(n-1)-2
Sn-S(n-1)=2an-2a(n-1)+n^2-(n-1)^2-3n+3(n-1)=2an-2a(n-1)+2n-4=an
an=2a(n-1)-2n+4 an-2n=2a(n-1)-4n+4=2a(n-1)-4(n-1)=2[a(n-1)-2(n-1)]
(an-2n)/[a(n-1)-2(n-1)]=2 所以数列{bn}是公比为2的等比数列.
(2) n=1时 有S1=a1=2a1+1-3-2=2a1-4 a1=4 b1=2 bn=2*2^(n-1)=2^n>0
Cn=1/(2^n+1)>0 C(n+1)=1/[2^(n+1)+1]
CnC(n+1)=1/{(2^n+1)[2^(n+1)+1]}=(1/2^n){1/(2^n+1)-1/[2^(n+1)+1]}
2^(n-1)CnC(n+1)=(1/2){1/(2^n+1)-1/[2^(n+1)+1]}
Tn=(1/2){1/3-1/5+1/5-1/9+.+1/(2^n+1)-1/[2^(n+1)+1]}=(1/2){1/3-1/[2^(n+1)+1]}
=1/6-(1/2){1/[2^(n+1)+1]}