Question

已知數(shù)列{a_n}滿足:a₁=1,a₂=2,2a_n=a_n-1+a_n+1(n≥2,n∈N^*),數(shù)列{b_n}滿足b₁=2,a_nb_n+1=2a_n+1b_n.
(1)求數(shù)列{a_n}的通項(xiàng)a_n;
(2)求證:數(shù)列為等比數(shù)列,并求數(shù)列{b_n}的通項(xiàng)公式.

Answer 1

(1) a_n= n    (2) b_n=n·2ⁿ

解析解:(1)∵2a_n=a_n-1+a_n+1,∴數(shù)列{a_n}為等差數(shù)列.
又a₁=1,a₂=2,所以d=a₂-a₁=2-1=1,
數(shù)列{a_n}的通項(xiàng)a_n=a₁+(n-1)d=1+(n-1)×1=n.
(2)∵a_n=n,∴nb_n+1=2(n+1)b_n,∴=2·,
所以數(shù)列是以=2為首項(xiàng),q=2為公比的等比數(shù)列,
∴=2×2^n-1,∴b_n=n·2ⁿ.