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