作业帮已知数列{an}满足a1=1/4,an=[a(n-1)]/[3a(n-1)+1] (n∈N,n≥2),求数列{1/an}的通项公式.
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![已知数列{an}满足a1=1/4,an=[a(n-1)]/[3a(n-1)+1] (n∈N,n≥2),求数列{1/an}的通项公式.](/uploads/image/z/542285-53-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2F4%2Can%3D%5Ba%28n-1%29%5D%2F%5B3a%28n-1%29%2B1%5D+%28n%E2%88%88N%2Cn%E2%89%A52%29%2C%E6%B1%82%E6%95%B0%E5%88%97%7B1%2Fan%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F.)
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已知数列{an}满足a1=1/4,an=[a(n-1)]/[3a(n-1)+1] (n∈N,n≥2),求数列{1/an}的通项公式.
已知数列{an}满足a1=1/4,an=[a(n-1)]/[3a(n-1)+1] (n∈N,n≥2),求数列{1/an}的通项公式.
已知数列{an}满足a1=1/4,an=[a(n-1)]/[3a(n-1)+1] (n∈N,n≥2),求数列{1/an}的通项公式.
取倒数
1/an=[3a(n-1)+1]/a(n-1)
1/an=3+1/a(n-1)
1/an-1/a(n-1)=3
所以1/an是等差数列,d=3
1/an=1/a1+(n-1)d=3n+1
an=1/(3n+1)