Question

已知數(shù)列{an}中,a1=1,前n項(xiàng)和為Sn且Sn+1=Sn+1,n∈N*.

(1)求數(shù)列{an}的通項(xiàng)公式.

(2)求數(shù)列{}的前n項(xiàng)和Tn.

 

Answer 1

(1) an=()n-1 (2) Tn=3[1-()n]

【解析】(1)由Sn+1=Sn+1,

得當(dāng)n≥2時(shí)Sn=Sn-1+1,

∴Sn+1-Sn=(Sn-Sn-1),

即an+1=an,∴=,

又a1=1,得S2=a1+1=a1+a2,∴a2=,

∴=,

∴數(shù)列{an}是首項(xiàng)為1,公比為的等比數(shù)列,

∴an=()n-1.

(2)∵數(shù)列{an}是首項(xiàng)為1,公比為的等比數(shù)列,

∴數(shù)列{}是首項(xiàng)為1,公比為的等比數(shù)列,

∴Tn==3[1-()n].