當(dāng)n=1時(shí)，a₁=S₁=2-1=1．
當(dāng)n≥2時(shí)，a_n=S_n-S_n-1=2ⁿ-1-（2^n-1-1）=2^n-1．
當(dāng)n=1時(shí)也成立．
∴an＝2n?1．
∴當(dāng)n≥2時(shí)，

a 2n

a 2n?1

=

(2n?1)2

(2n?2)2

=4．
∴數(shù)列{

a

2n

}是等比數(shù)列，首項(xiàng)為

a

21

=1，公比為4．
∴

a

21

+

a

22

+

a

23

+…+

a

2n

=

4n?1

4?1

=

1

3

(4n?1)．
故選：D．