等比数列{an},Sn=2^n-1,则a1^2+a2^2+...+a^n=?

Sn=2^n -1,
S(n-1)=2^(n-1) -1,
an
=Sn-S(n-1)
=2^n -1-(2^(n-1) -1)
=2^n-2^(n-1)
=2^(n-1)
an^2=2^(2n-2)=(4^n)/4,
a(n+1)^2=4^(n+1)/4,
a(n+1)^2/an^2=4
an^2是以a1^2=1为首项,4为公比的等比数列；
S=(1-4^n)/(1-4)=(4^n-1)/3.