题目

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

提问时间：2020-07-26

答案

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-22a(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)...