等比数列{a}的前n项和为Sn=2^n-1,则a1^2+a2^2+.an^2=
问题描述：
等比数列{a}的前n项和为Sn=2^n-1,则a1^2+a2^2+.an^2=
答
Sn=2^n-1
S(n-1)=2^(n-1)-1=2^n/2-1
Sn-S(n-1)=an=2^n-1-2^n/2+1=2^n/2=2^(n-1)
an^2=(2^(n-1))^2=2^(2n-2)
a(n-1)^2=2^(2(n-1)-2)=2^(2n-2-2)=2^(2n-2)*2^(-2)
an^2/a(n-1)^2=2^(2n-2)/2^(2n-2)*2^(-2)=2^2=4
所以数列(an^2)也是等比数列,公比是4
a1^2=2^(2*1-2)=1
则
a1^2+a2^2+.an^2
=1*(4^n-1)/(4-1)
=(4^n-1)/3