Question

设各项均为正数的数列{a_n}和{b_n}满足,,成等比数列,lgb_n,lga_n+1,lgb_n+1成等差数列,且a₁=1,b₁=2,a₂=3,求a_n,b_n.

Answer 1

分析：由等比、等差中项的性质得a_n+1=,递推出a_n=(n≥2).

解：∵,, 成等比数列,

∴()²=·,即2b_n=a_n+a_n+1.                                 ①

    又∵lgb_n,lga_n+1,lgb_n+1成等差数列,

∴2lga_n+1=lgb_n+lgb_n+1,即a_n+1²=b_n·b_n+1.

∴a_n+1=.                                                            ②

∴a_n=(n≥2).                                                      ③

    把②③代入①得2b_n=(n≥2),

∴=(n≥2).

∴数列{}为等差数列.

∵b₁=2,a₂=3,=b₁·b₂,

∴b₂=,=+(n-1)()

=(n+1)(n=1也成立).

∴b_n=,

a_n===(n≥2).

    又当n=1时,a₁=1=,∴a_n=.