Question

在数列{an}中,a1=1,且对任意K∈N*,a(2k-1),a(2k),a(2k+1)成等比数列,其公比 根号下[（k

Answer 1

a(2k+1)/a(2k-1)=(k+1)/k

a(2k-1)/a(2k-3)=k/(k-1)...

a3/a1=2/1

左右累乘

a(2k+1)/a1=[2×3×...×(k+1)]/(1×2×3×...×k)=k+1

a(2k+1)=a1(k+1)=k+1

(2011-1)/2=1005 k=1005

a2011=1005+1=1006