数列{an}的前n项和为Sn,存在常数ABC,使得an+Sn=An^2+Bn+C对任意正整数都成立若A=-1/2,B=-3/2,C=1,设bn=an+n,数列{nbn}的前n项的和为Tn,求Tn

数列{an}的前n项和为Sn,存在常数ABC,使得an+Sn=An^2+Bn+C对任意正整数都成立若A=-1/2,B=-3/2,C=1,设bn=an+n,数列{nbn}的前n项的和为Tn,求Tn
数列{an}的前n项和为Sn,存在常数ABC,使得an+Sn=An^2+Bn+C对任意正整数都成立
若A=-1/2,B=-3/2,C=1,设bn=an+n,数列{nbn}的前n项的和为Tn,求Tn

数列{an}的前n项和为Sn,存在常数ABC,使得an+Sn=An^2+Bn+C对任意正整数都成立若A=-1/2,B=-3/2,C=1,设bn=an+n,数列{nbn}的前n项的和为Tn,求Tn
a(1)+s(1)=2a(1)=A+B+C=-1, a(1)=-1/2.
a(n+1)+s(n+1)-[a(n)+s(n)]=a(n+1)-a(n)+a(n+1)=A(2n+1)+B=2An+A+B=-n-2
2a(n+1)-a(n)=-n-2,
2[a(n+1)+(n+1)] - [a(n)+n] = -n-2+2n+2 - n =0,
2b(n+1)-b(n) = 0,
b(n+1)=(1/2)b(n),
b(1)=a(1)+1=-1/2+1=1/2.
b(n)=(1/2)^(n) = 1/2^n,
t(n)=1/2 + 2/2^2 + 3/2^3 + ... + (n-1)/2^(n-1) + n/2^n,
2t(n)=1/1 + 2/2 + 3/2^2 + ... + (n-1)/2^(n-2) + n/2^(n-1),
t(n)=2t(n)-t(n)=1/1 + 1/2 + 1/2^2 + ... + 1/2^(n-1) - n/2^n
=[1-(1/2)^(n)]/[1-1/2] - n/2^n
=2[1-1/2^n] - n/2^n
=2 - (n+2)/2^n