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设数列{xn}满足logaxn+1=1+logaxn(a>0,a≠1),若x1+x2+…+x100=100,则x101+

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设数列{xn}满足logaxn+1=1+logaxn(a>0,a≠1),若x1+x2+…+x100=100,则x101+x102+…+x200=______.
∵logaxn+1=1+logaxn,∴logaxn+1-logaxn=1,

log
xn+1
xna=1,则
xn+1
xn=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