Question

已知数列{a n }满足:a 1 =1,a 2 =2,2a n =a n-1 +a n+1 (n≥2,n∈N * ),数列{b n }满足b 1 =2,a n b n+1

已知数列{a n }满足:a 1 =1,a 2 =2,2a n =a n-1 +a n+1 (n≥2,n∈N * ),数列{b n }满足b 1 =2,a n b n+1 =2a n+1 b 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

(1) a n = n    (2) b n =n·2 n

解:(1)∵2a n =a n-1 +a n+1 ,∴数列{a n }为等差数列. 又a 1 =1,a 2 =2,所以d=a 2 -a 1 =2-1=1, 数列{a n }的通项a n =a 1 +(n-1)d=1+(n-1)×1=n. (2)∵a n =n,∴nb n+1 =2(n+1)b n ,∴ =2· , 所以数列 是以 =2为首项,q=2为公比的等比数列, ∴ =2×2 n-1 ,∴b n =n·2 n .