Question

已知數(shù)列{a_n}的前n項(xiàng)和為S_n,a₁=1,S_n=2a_n+1,則S_n等于(　　)

A．2n-1

B．n-1

C．n-1

D．

Answer 1

B

法一　由S_n=2a_n+1=2(S_n+1-S_n)可知,
3S_n=2S_n+1,即S_n+1=S_n,
∴數(shù)列{S_n}是首項(xiàng)為S₁=1,公比為的等比數(shù)列,
∴S_n=^n-1.故選B.
法二　由S_n=2a_n+1�、倏芍猘₂=S₁=,
當(dāng)n≥2時(shí),S_n-1=2a_n,�、�
∴①-②并化簡得a_n+1=a_n(n≥2),
即{a_n}從第二項(xiàng)起是首項(xiàng)為,公比為的等比數(shù)列,
∴S_n=a₁+=1+^n-1-1=^n-1(n≥2),當(dāng)n=1時(shí),滿足上式.
故選B.
法三　特殊值法,由S_n=2a_n+1及a₁=1,
可得a₂=S₁=,
∴當(dāng)n=2時(shí),S₂=a₁+a₂=1+=,觀察四個(gè)選項(xiàng)得B正確.故選B.