Question

数列{an}的前项n的和为Sn,存在常数A、B、C，使得an+Sn=An^2+Bn+C对任意正整数n都成立。(1)若数列{an}为等差

数列{an}的前项n的和为Sn,存在常数A、B、C，使得an+Sn=An^2+Bn+C对任意正整数n都成立。(1)若数列{an}为等差数列，求证：3A-B+C＝0; （2）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;

Answer 1

1.不妨设公差为d 首项为a1 an=a1+(n-1)d sn=a1n +n(n-1)d/2 an+sn=dn^2/2 +(a1+0.5d)n+a1-d
即A=0.5d B=a1+0.5d C=a1-d vvv3A-B+C=1.5d-a1-0.5+a1-d=0成立
2.对n=1 2a1=-1 a1=-1/2 b1=1/2 设bn的前n项和为Rn 则Rn=Sn+0.5n^2+0.5n 而an+sn=-0.5n^2-1.5n+1 an+n+sn+0.5n^2=-0.5n+1 bn+Rn=1 对于n≥2 b(n-1)+R(n-1)=1 则bn-b(n-1)+bn=0
bn=0.5b(n-1) 所以bn是以b1=0.5为首项 0.5为公比的等比数列 bn=0.5^n
Tn=1*0.5+2*0.5^2+3*0.5^3+……+n*0.5^n
0.5Tn= 1*0.5^2+2*0.5^3+……+(n-1)*0.5^n +n*0.5^(n+1) 上下相减得
0.5Tn=0.5+0.5^2+……+0.5^n- n0.5^(n+1)=1-0.5^n-n0.5^(n+1)
Tn=2-(2+n)0.5^n