数列{an}中,a1=1sn是{an}的前n项和,当n大于等于2是sn=an[1-2/sn]求证{1/sn}是等差数列第二问 tn=s1×s2+s2×s3+.+sn×s(n+1 ) 求tn
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![数列{an}中,a1=1sn是{an}的前n项和,当n大于等于2是sn=an[1-2/sn]求证{1/sn}是等差数列第二问 tn=s1×s2+s2×s3+.+sn×s(n+1 ) 求tn](/uploads/image/z/1807239-39-9.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D1sn%E6%98%AF%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%2C%E5%BD%93n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%E6%98%AFsn%3Dan%5B1-2%2Fsn%5D%E6%B1%82%E8%AF%81%7B1%2Fsn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E7%AC%AC%E4%BA%8C%E9%97%AE+tn%3Ds1%C3%97s2%2Bs2%C3%97s3%2B.%2Bsn%C3%97s%EF%BC%88n%2B1+%29+%E6%B1%82tn)
数列{an}中,a1=1sn是{an}的前n项和,当n大于等于2是sn=an[1-2/sn]求证{1/sn}是等差数列第二问 tn=s1×s2+s2×s3+.+sn×s(n+1 ) 求tn
数列{an}中,a1=1sn是{an}的前n项和,当n大于等于2是sn=an[1-2/sn]求证{1/sn}是等差数列
第二问 tn=s1×s2+s2×s3+.+sn×s(n+1 ) 求tn
数列{an}中,a1=1sn是{an}的前n项和,当n大于等于2是sn=an[1-2/sn]求证{1/sn}是等差数列第二问 tn=s1×s2+s2×s3+.+sn×s(n+1 ) 求tn
1/Sn=(n+1)/2 则Sn=2/(n+1)
S1xS2+S2xS3+.+SnxS(n+1)=4(1/2 x 1/3 +1/3 x 1/4 +.)
=4(1/2 - 1/3 +1/3 -1/4 +.)
=4(1/2 -1/(n+2))
n≥2时
an=Sn-Sn-1
Sn=(Sn-Sn-1)(1-2/Sn)
化简得1/Sn=1/2 + 1/Sn-1
所以{1/sn}是等差数列 公差为1/2
解得Sn=2/(n+1)
tn=2/2×2/3+2/3×2/4+......2/n×2/(n+1)
∵1/n×1/(n+1) =1/n-1/(n+1)
∴tn=2/2-2...
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n≥2时
an=Sn-Sn-1
Sn=(Sn-Sn-1)(1-2/Sn)
化简得1/Sn=1/2 + 1/Sn-1
所以{1/sn}是等差数列 公差为1/2
解得Sn=2/(n+1)
tn=2/2×2/3+2/3×2/4+......2/n×2/(n+1)
∵1/n×1/(n+1) =1/n-1/(n+1)
∴tn=2/2-2/3+2/3-2/4+......2/n-2/(n+1)
tn=1-2/(n+1)
tn=(n-1)/(n+1)
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