Question

已知數(shù)列{a_n}的前n項(xiàng)和S_n=2ⁿ-1，則此數(shù)列奇數(shù)項(xiàng)的前n項(xiàng)和為（　�。�

A． 2n+1-1 3

B． 2n+1-2 3

C． 22n-1 3

D． 22n-2 3

Answer 1

當(dāng)n=1時(shí)，a₁=S₁=2¹-1=1，
當(dāng)n≥2時(shí)，a_n=Sn-Sn-1=2ⁿ-1-（2^n-1-1）=2•2^n-1-2^n-1=2^n-1，對(duì)n=1也適合
∴a_n=2^n-1，
∴數(shù)列{a_n}是等比數(shù)列，此數(shù)列奇數(shù)項(xiàng)也構(gòu)成等比數(shù)列，且首項(xiàng)為1，公比為4．
∴此數(shù)列奇數(shù)項(xiàng)的前n項(xiàng)和為

1•(1-4n)

1-4

=

4n-1

3

=

22n-1

3

故選C