已知{an}是公差为d的等差数列,bn=1/n(a1+a2+……an),数列{an}、{bn}的前n项和分别是Sn、Tn
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已知{an}是公差为d的等差数列,bn=1/n(a1+a2+……an),数列{an}、{bn}的前n项和分别是Sn、Tn,若S25-T25
![已知{an}是公差为d的等差数列,bn=1/n(a1+a2+……an),数列{an}、{bn}的前n项和分别是Sn、Tn](/uploads/image/z/821678-14-8.jpg?t=%E5%B7%B2%E7%9F%A5%7Ban%7D%E6%98%AF%E5%85%AC%E5%B7%AE%E4%B8%BAd%E7%9A%84%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2Cbn%3D1%2Fn%28a1%2Ba2%2B%E2%80%A6%E2%80%A6an%29%2C%E6%95%B0%E5%88%97%7Ban%7D%E3%80%81%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E5%88%86%E5%88%AB%E6%98%AFSn%E3%80%81Tn)
题目不完整啊!
bn=1/n(a1+a2+...+an)
=1/n *(a1+an)n/2
=(a1+an)/2
=a1+(n-1)d/2
设cn=an-bn=(1-n)d/2=d/2-nd/2
S25-T25=c1+c2+...+c25
=25*d/2-(1+2+..+25)d/2
=-(1+2+...+24)d/2
=-150d∈(0,1),
-1/150
bn=1/n(a1+a2+...+an)
=1/n *(a1+an)n/2
=(a1+an)/2
=a1+(n-1)d/2
设cn=an-bn=(1-n)d/2=d/2-nd/2
S25-T25=c1+c2+...+c25
=25*d/2-(1+2+..+25)d/2
=-(1+2+...+24)d/2
=-150d∈(0,1),
-1/150
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