数列{an}满足a1=1/2,an+1=1/2-an(n属于正整数)设bn=1/1-an,证明{bn}是等差数列,并求bn和an各位哥哥姐姐们谁会啊?
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![数列{an}满足a1=1/2,an+1=1/2-an(n属于正整数)设bn=1/1-an,证明{bn}是等差数列,并求bn和an各位哥哥姐姐们谁会啊?](/uploads/image/z/400405-13-5.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a1%3D1%2F2%2Can%2B1%3D1%2F2-an%EF%BC%88n%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89%E8%AE%BEbn%3D1%2F1-an%2C%E8%AF%81%E6%98%8E%7Bbn%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82bn%E5%92%8Can%E5%90%84%E4%BD%8D%E5%93%A5%E5%93%A5%E5%A7%90%E5%A7%90%E4%BB%AC%E8%B0%81%E4%BC%9A%E5%95%8A%3F)
数列{an}满足a1=1/2,an+1=1/2-an(n属于正整数)设bn=1/1-an,证明{bn}是等差数列,并求bn和an各位哥哥姐姐们谁会啊?
数列{an}满足a1=1/2,an+1=1/2-an(n属于正整数)
设bn=1/1-an,证明{bn}是等差数列,并求bn和an
各位哥哥姐姐们谁会啊?
数列{an}满足a1=1/2,an+1=1/2-an(n属于正整数)设bn=1/1-an,证明{bn}是等差数列,并求bn和an各位哥哥姐姐们谁会啊?
an+1=1/2-an两边减1得:
(an+1)-1=1/(2-an)-1,再取倒数整理得:
1/[(an+1)-1]=-1+1/(an-1),
所以,数列{1/[(an)-1]}是首项-2,公差-1的等差数列,∴1/[(an)-1]=-n-1,
解得:an=n/(n+1).
∵bn=1/1-an,代入an整理得:bn=n+1,故bn+1=n+2,(bn+1)-bn=n+2-n-1=1,∴{bn}是等差数列.
下标易混,注意识别.自己写一写更清楚.
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