数列{an}中,a1=1,n≥2时,其前n项的和Sn满足Sn2=an(Sn-1/2) (1)求Sn的表达式; (2)设bn=Sn2n+1,数列{bn}的前n项和为Tn,求limn→∞Tn.
问题描述:
数列{an}中,a1=1,n≥2时,其前n项的和Sn满足Sn2=an(Sn-
)1 2
(1)求Sn的表达式;
(2)设bn=
,数列{bn}的前n项和为Tn,求Sn 2n+1
Tn. lim n→∞
答
(1)n≥2,sn2=(sn-sn-1)(sn-12)∴sn=sn−12sn−1+1即1sn-1sn−1=2(n≥2)∴1sn=2n-1故sn=12n−1(2)bn=sn2n+1=1(2n+1)(2n−1)=12(12n−1-12n+1)Tn=12(1-13+13-15+15-17+…+12n−1-12n+1)+=12(1-12n+1)...