Question

已知数列{an}的前n项和为Sn满足Sn=2^n-1,则数列{an^2}的前几项为（）

要步骤！

Answer 1

Sn=2^n-1
S(n-1)=2^(n-1)-1
Sn-S(n-1)=an
=2^n-1 -2^(n-1)+1
=2*2^(n-1)-2^(n-1)
=2^(n-1)

an^2=[2^(n-1)]^2
=2^(2n-2)
∴
a1^2=1
a2^2=2^2=4
a3^3=2^4=16
a4^2=2^6=64
a5^2=2^8=256
..............
an^2=2^(2n-2)

Answer 2

Sn=2^n-1,S1=1
Sn-1=2^(n-1)-1
∴an=Sn-S(n-1)=2^n-2^(n-1)=2^(n-1)
∴bn=an^2=(2^(n-1))^2=4^(n-1)
∴数列{an^2}的前几项为:1,4,16,64……4^(n-1)
前n项和Tn=(4^n-1)/3