Question

已知数列{an}满足a1=2，an+1=（3an-2）/（2an-1）一题一题打、快点

（1）写出a2，a3的值（2）求证{1/（an-1）}为等差数列（3）设bn=an/[2n*（2n+1）]，Sn为数列{bn}的前n项和，求证Sn<1/2

Answer 1

(1)因为A1=2
则A2=(3A1-2)/(2A1-1)=(3×2-2)/(2×2-1)=4/3
A3=（3A2-2)/(2A2-1)=(3×4/3-2)/(2×4/3-1)=6/5

(2)设Bn=1/(An-1) ==>An=(bn+1)/bn
代入可得(b[n+1]+1)/bn=[3(bn+1)/bn-2]/[2(bn+1)/bn-1]
化简可得b[n+1]-bn=2
所以数列bn是以b1=1/(a1-1)=1为首项,2为公差的等差数列
所以bn=1+2(n-1)=2n-1
代入可得an=2n/(2n-1)

(3)bn=2n/(2n-1)÷2n/(2n+1)=1/(2n-1)(2n+1)=1/2[1/(2n-1)-1(2n+1)]
所以Sn=1/2[1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=1/2(1-1/(2n+1)
<1/2