等比数列｛an｝中，已知对任意自然数n，a1+a2+a3+……+an＝2^n-1，则a1^2+a2

a1+a2+a3+……+an=Sn=2^n-1 S(n-1)=2^(n-1)-1 Sn-S(n-1)=an =2^n-2^(n-1) =2^(n-1) an^2=2^2(n-1)=2^(2n-2) a(n-1)^2=2^2(n-2)=2^(2n-4) an^2/a(n-1)^2=2^(2n-2) / 2^(2n-4)=2^2=4 ∴数列{an^2}是等比数列，公比q=4 a1^2=1 a1^2+a2^2+..........an^2 =a1(q^n-1)(q-1