数列{bn}满足b1=2,4/｛b(n+1)bn｝-6/b(n+1)+3/bn=o（n≥1）1、求｛bn｝的通项公式.2、记bn=1/(an-1/2)求数列｛anbn}的前n项和.

题目

数列{bn}满足b1=2,4/｛b(n+1)bn｝-6/b(n+1)+3/bn=o（n≥1）1、求｛bn｝的通项公式.2、记bn=1/(an-1/2)求数列｛anbn}的前n项和.

答案

1.4/｛b(n+1)bn｝-6/b(n+1)+3/bn=o
b(n+1) = 2bn - 4/3
b(n+1) - 4/3 = 2(bn - 4/3)
所以{ bn - 4/3}是首项为b1-4/3 = 2/3,公比为2的等比数列
bn - 4/3 = 2/3 * 2^(n-1) = 1/3*2^n
bn = 2^n /3 + 4/3
2.bn=1/(an-1/2)
所以 anbn = bn/2 + 1
所以 S(anbn) = S(bn)/2 + n
= 1/2*[1/3 *(2^1 + 2^2 + ……2^n) + 4n/3] + n
=(2^n-1)/3 + 2n/3 + n
=(2^n-1)/3 + 5n/3