设函数f(x)=x/(2x+1),数列{an}满足an=f(an-1),且a1=f(2)
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设函数f(x)=x/(2x+1),数列{an}满足an=f(an-1),且a1=f(2)
(1)求证{1/an}是等差数列
(2)求数列{an}的通项公式
(1)求证{1/an}是等差数列
(2)求数列{an}的通项公式
令1/an = bn
an = an-1 / (2an-1 + 1)
1/an = 2 + 1/an-1
bn = 2 + bn-1
则bn是以2为公差的等差数列
即1/an是等差数列
a1=f(2)=2/5
b1 = 1/a1 = 5/2
则bn = 5/2 + 2(n-1)
= 2n + 1/2
即1/an = 2n + 1/2
则an = 1/(2n + 1/2) = 2/(4n + 1)
an = an-1 / (2an-1 + 1)
1/an = 2 + 1/an-1
bn = 2 + bn-1
则bn是以2为公差的等差数列
即1/an是等差数列
a1=f(2)=2/5
b1 = 1/a1 = 5/2
则bn = 5/2 + 2(n-1)
= 2n + 1/2
即1/an = 2n + 1/2
则an = 1/(2n + 1/2) = 2/(4n + 1)
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