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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

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2 | 2012-8-10 15:39:25 | 显示全部楼层 |阅读模式

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) +(1/2-1/3)+...+ (1/n-1/(n+1)) ]
=(1/4)[ 1- 1/(n+1) ]
= n/[4(n+1)]...

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