摘要:11.设正项等比数列{an}的前n项和为Sn.已知a3=4.a4a5a6=212. (1)求首项a1和公比q的值, (2)若Sn=210-1.求n的值. 解:(1)设等比数列{an}的公比为q(q>0).则由题设有a4a5a6=a=212⇒a5=24=16. ∴=q2=4⇒q=2.代入a3=a1q2=4.解得a1=1. (2)由Sn=210-1.得Sn==2n-1=210-1.∴2n=210.∴n=10.
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