【高一数学】数列{an}前n项和Sn=-n^2+9n,bn=|an|设bn前n项和为Tn,求Tn
【高一数学】数列{an}前n项和Sn=-n^2+9n,bn=|an|设bn前n项和为Tn,求Tn
数学人气:579 ℃时间:2019-08-21 13:46:32
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解由Sn=-n^2+9n当n≥2时,Sn=-(n-1)^2+9(n-1)两式相减得an=-2n+10(n≥2)当n=1时,a1=S1=-1²+9=8对an=-2n+10成立即数列{an}的通项公式an=-2n+10令an≥0,即n≤5,即bn=an=-2n+10即an<0,即n>5,即bn=-an=2n-10...
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