数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010

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数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010
数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010

数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010
a(n+1)=an/(2an+1)
2a(n+1)an+a(n+1)=an
an-a(n+1)=2a(n+1)an
等式两边同除以a(n+1)an
1/a(n+1)-1/an=2,为定值.
1/a1=1/1=1
数列{1/an}是以1为首项,2为公差的等差数列.
通项公式为1/an=1+2(n-1)=2n-1
an=1/(2n-1)
所以 a(2010)=1/4019

a(1)=1
a(2)=1/3
a(3)=1/5
a(4)=1/7
猜想 a(n)=1/(2n-1)
所以 a(2010)=1/4019
严谨的做法
两边取倒数 1/a(n+1)=2+1/a(n)
所以{1/a(n)}是等差数列，公差为2，首项为1
1/an=1+2(n-1)=2n-1
所以 a(n)=1/(2n-1)
所以 a(2010)=1/4019

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