Question

等比数列{an}中,n为正整数,a1+a2+.+an=2^n-1,求a1^2+a2^2+...+an^2=?

Answer 1

a1+a2+.+an=2^n-1

a1+a2+.+a(n-1)=2^(n-1)-1

相减得an=2^(n-1)

an^2=4^(n-1)

a1^2+a2^2+a3^2+……an^2

=1*(1-4^n)/(1-4)

=(4^n-1)/3