在数列{an}中,已知a1=3/5,an*a(n-1)+1=2a(n-1)(n>=2,n∈N),数列{bn}满足:bn=1/(an-1)(n∈N*)(1)求证:{bn}是等差数列;(2)求数列{an}中得最大项与最小项,并说明理由.
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![在数列{an}中,已知a1=3/5,an*a(n-1)+1=2a(n-1)(n>=2,n∈N),数列{bn}满足:bn=1/(an-1)(n∈N*)(1)求证:{bn}是等差数列;(2)求数列{an}中得最大项与最小项,并说明理由.](/uploads/image/z/2722305-57-5.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5a1%3D3%2F5%2Can%2Aa%28n-1%29%2B1%3D2a%28n-1%29%28n%3E%3D2%2Cn%E2%88%88N%29%2C%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Abn%3D1%2F%28an-1%29%28n%E2%88%88N%2A%29%281%29%E6%B1%82%E8%AF%81%EF%BC%9A%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%9B%282%29%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%E5%BE%97%E6%9C%80%E5%A4%A7%E9%A1%B9%E4%B8%8E%E6%9C%80%E5%B0%8F%E9%A1%B9%2C%E5%B9%B6%E8%AF%B4%E6%98%8E%E7%90%86%E7%94%B1.)
在数列{an}中,已知a1=3/5,an*a(n-1)+1=2a(n-1)(n>=2,n∈N),数列{bn}满足:bn=1/(an-1)(n∈N*)(1)求证:{bn}是等差数列;(2)求数列{an}中得最大项与最小项,并说明理由.
在数列{an}中,已知a1=3/5,an*a(n-1)+1=2a(n-1)(n>=2,n∈N),数列{bn}满足:bn=1/(an-1)(n∈N*)
(1)求证:{bn}是等差数列;
(2)求数列{an}中得最大项与最小项,并说明理由.
在数列{an}中,已知a1=3/5,an*a(n-1)+1=2a(n-1)(n>=2,n∈N),数列{bn}满足:bn=1/(an-1)(n∈N*)(1)求证:{bn}是等差数列;(2)求数列{an}中得最大项与最小项,并说明理由.
(1)an*a(n-1)+1=2a(n-1)
an=[2a(n-1)-1]/a(n-1)
an -1= [2a(n-1)-1]/a(n-1) -1= [a(n-1)-1]/a(n-1)
1/(an -1) = a(n-1)/[a(n-1)-1] = {[a(n-1)-1]+1} /[a(n-1)-1]
=1+ 1/[a(n-1) -1]
1/(an -1) -1/[a(n-1) -1] =1
即 bn - b(n-1)= 1
所以bn 是等差数列.
(2) bn =b1 +1*(n-1)= 1/(a1 -1) +n-1=1/(-2/5) +n-1= n-7/2,n>=2
an =1/bn +1 =1/(n-7/2) +1= 1+ 2/(2n-7)
当2n-7>0,且2n-7为最小时,an有最大值,
2n-7>0,n>7/2,n=4时,2n-7>0且最小,此时a4最大,a4=1+2/(2*4-7)=3
当2n-7