已知数列{an}的前n项和Sn,且满足an+2SnSn-1=0(n2),a1=1/2.求Tn=s1s2+s2s3+...+sns

an + 2SnS(n-1) =0Sn -S(n-1) =-2SnS(n-1)1/Sn -1/S(n-1) =21/Sn -1/S1 =2(n-1)1/Sn= 2nSn = 1/(2n)Sn.S(n+1) = 1/(2n) .1/(2n+2)= (1/4) ( 1/n -1/(n+1) )Tn = S1S2 + S2S3 +..+ SnS(n+1)= (1/4)[ (1/1-1/2) +(1/2-...