等比数列{an}中,若Sn=2的n次方+C,则a1²+a2²+...+an²=?

a1=s1=2^1+C=2+C
an=sn-s(n-1)
=2^n+C-2^(n-1)-C
=2^(n-1)
a1=2^(1-1)=2^0=1
2+C=1
C=-1
(a1)^2+(a2)^2+(a3)^2+.+(an)^2
=2^0+2^2+2^4+.+2^(2n-2)
=(1-4^n)/(1-4)
=(4^n-1)/3