Question

已知数列{a n }的前n项和为S n ,a 1 =1,S n =2a n+1 ,则S n 等于(　　) A．2 n-1 B． n-1 C．

已知数列{a n }的前n项和为S n ,a 1 =1,S n =2a n+1 ,则S n 等于(　　) A．2 n-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 , ∴数列{S n }是首项为S 1 =1,公比为 的等比数列, ∴S n = n-1 .故选B. 法二　由S n =2a n+1 　①可知a 2 = S 1 = , 当n≥2时,S n-1 =2a n ,　② ∴①-②并化简得a n+1 = a n (n≥2), 即{a n }从第二项起是首项为 ,公比为 的等比数列, ∴S n =a 1 + =1+ n-1 -1= n-1 (n≥2),当n=1时,满足上式. 故选B. 法三　特殊值法,由S n =2a n+1 及a 1 =1, 可得a 2 = S 1 = , ∴当n=2时,S 2 =a 1 +a 2 =1+ = ,观察四个选项得B正确.故选B.