Question

設(shè)Sn為數(shù)列{an}的前n項(xiàng)和,已知a1≠0,2an-a1=S1·Sn,n∈N*.

(1)求a1,a2,并求數(shù)列{an}的通項(xiàng)公式;

(2)求數(shù)列{nan}的前n項(xiàng)和.

Answer 1

解:(1)令n=1,得2a1-a1=,即a1=.

因?yàn)閍1≠0,所以a1=1.

令n=2,得2a2-1=S2=1+a2,解得a2=2.

當(dāng)n≥2時(shí),由2an-1=Sn,2an-1-1=Sn-1兩式相減,

得2an-2an-1=an,即an=2an-1.

于是數(shù)列{an}是首項(xiàng)為1,公比為2的等比數(shù)列.

因此,an=2n-1.所以數(shù)列{an}的通項(xiàng)公式為an=2n-1.

(2)由(1)知,nan=n·2n-1.

記數(shù)列{n·2n-1}的前n項(xiàng)和為Bn,

于是Bn=1+2×2+3×22+…+n×2n-1,①

2Bn=1×2+2×22+3×23+…+n×2n.②

①-②,得-Bn=1+2+22+…+2n-1-n·2n=2n-1-n·2n.

從而Bn=1+(n-1)·2n.