设数列{xn}满足logaxn+1=1+logaxn(a>0,a≠1),若x1+x2+…+x100=100,则x101+x102+…+x200=______.
题型:填空题难度:简单来源:不详
设数列{xn}满足logaxn+1=1+logaxn(a>0,a≠1),若x1+x2+…+x100=100,则x101+x102+…+x200=______. |
答案
∵logaxn+1=1+logaxn,∴logaxn+1-logaxn=1, ∴=1,则=a, ∴数列{xn}是以a为公比的等比数列, ∵x1+x2+…+x100=100,∴x101+x102+…+x200=a100x1+a100x2+…a100x100 =a100(x1+x2+…+x100)=100a100, 故答案为:100a100. |
举一反三
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