Question

设数列{an}满足a1+a2+a3+……+an=2^n-1对任意正整数n都成立,则1/a1+1/a3+1/a5+……+1

Answer 1

a1=1 =2^0 a1＋a2=2²－1=3 ∴a2=2=2¹ a1＋a2＋a3=2³-1=7 ∴a3=4=2² 依次类推 an=2^﹙n－1﹚

∴ 1/a1+1/a3+1/a5+……+1/a2n-1+1/a2n+1=1＋1／2²＋1／2^4＋1/2^2n

公比是1／2²=¼

∴Sn=[a1*(1-q^n)]/(1-q)=[1-﹙¼﹚^n]／﹙1－¼﹚