Question

在等比数列{a n}中,已知对任意正整数n,a1+a2+……+an=2^n-1,则a1^2+a2^2+……+an^2=?

Answer 1

a2^2/a1^2=q^2

a1+a2+……+an=2^n-1

Sn=a1(1-q^n)/(1-q)=2^n-1

由此可得：a1=1,q=2

设：Tn=a1^2+a2^2+……+an^2

=a1^2(1-q^2n)/(1-q^2)

=4(4^n-1)/3