Question

已知数列{an}满足a1=4,an=4-4/an-1

已知数列{an}满足a1=4,an=4-4/a（n-1）(n≥2)，令bn=1/ an-2. 1、求证：数列{bn}是等差数列 2、求数列{an}通项

Answer 1

（1）证明：
an=4-4/a(n-1)
a(n)-2=2- 4/a(n-1)
b(n)=1/（a(n)-2）=1/[2-4/a(n-1)]=a(n-1)/[2a(n-1)-4]=a(n-1)/2[a(n-1)-2]=1/2 +1/[a(n-1)-2]
bn-b(n-1)=[1/(an-2)]=1/2为定值
所以数列{bn}是等差数列
(2)解：
{bn}首项是1/(4-2)=1/2,公差是1/2
bn＝1/2+(1/2)*(n-1)=n/2
1/（an-2）＝n／2
an-2=2/n
an=2+2/n

Answer 2

an=4-4/a(n-1)
an-2
=2-4/a(n-1)
=2{[a(n-1)-2]/a(n-1)}
1/(an-2)=1/2+1/[a(n-1)-2]、
1/（a1-2）=1/2
bn=1/(an-2)=1/2+b(n-1)
bn-b(n-1)=1/2
所以数列{bn}是以1/2为首项，以1/2为公关的等差数列，
bn=n/2
1/(an-2)=n/2
得an=(2/n)+2
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Answer 3

(1)
an=4-4/a(n-1)

an -2 = 2(a(n-1) - 2)/a(n-1)
1/[an -2 ] = a(n-1)/[ 2(a(n-1) - 2) ]
= 1/2 + 1/(a(n-1) - 2)
1/[an -2 ] -1/(a(n-1) - 2) = 1/2

bn =1/(an - 2)
{bn}是等差数列, d=1/2

(2)
1/[an -2 ] -1/(a(n-1) - 2) = 1/2
1/[an -2 ] -1/(a1 - 2) = (n-1)/2
1/[an -2 ] =n/2
an = 2+ 2/n