(2014•呼和浩特一模)数列{an},已知对任意正整数n,a1+a2+a3+…+an=2n-1,则a12+a22+a32+…+an2 等于(  ) A.(2n-1)2 B.13(2n−1) C.13(4n−1) D.4n-1

问题描述:

(2014•呼和浩特一模)数列{an},已知对任意正整数n,a1+a2+a3+…+an=2n-1,则a12+a22+a32+…+an2 等于(  )
A. (2n-1)2
B.

1
3
(2n−1)
C.
1
3
(4n−1)

D. 4n-1

∵a1+a2+a3+…+an=2n-1…①
∴a1+a2+a3+…+an-1=2n-1-1…②,
①-②得an=2n-1
∴an2=22n-2
∴数列{an2}是以1为首项,4为公比的等比数列,
∴a12+a22+a32+…+an2=

1−4n
1−4
=
1
3
(4n−1)

故选C.