Question

已知数列{a_n}满足:a₁=1,a₂=2,2a_n=a_n-1+a_n+1(n≥2,n∈N^*),数列{b_n}满足b₁=2,a_nb_n+1=2a_n+1b_n.

(1)求数列{a_n}的通项a_n;

(2)求证:数列为等比数列,并求数列{b_n}的通项公式.

Answer

(1) an= n (2) bn=n·2n

【解析】

解:(1)∵2an=an-1+an+1,∴数列{an}为等差数列.

又a1=1,a2=2,所以d=a2-a1=2-1=1,

数列{an}的通项an=a1+(n-1)d=1+(n-1)×1=n.

(2)∵an=n,∴nbn+1=2(n+1)bn,∴=2·,

所以数列是以=2为首项,q=2为公比的等比数列,

∴=2×2n-1,∴bn=n·2n.