在等比数列中{an}中,已知对于任意的n属于n+,有a1+a2+a3+……+an=2^n-1,则a1^2+a2^2+a3^2+……+an^2=

Sn=a1(q^n-1)/(q-1)根据题意,即等式a1(q^n-1)/(q-1)=2^n-1恒成立.[a1/(q-1)]q^n-[a1/(q-1)]=2^n-1a1/(q-1)=1q=2解得a1=1 q=2设数列{bn} b1=a1^2=1bn=an^2=[a1q^(n-1)]^2=2^[2(n-1)]=4^(n-1)数列{bn}是以1为首项,4为...