已知函数f(x)=(x+3)/(x+1) (x不等于-1),设数列{an}满足a1=1,an+1(n+1是下标)=f(a
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已知函数f(x)=(x+3)/(x+1) (x不等于-1),设数列{an}满足a1=1,an+1(n+1是下标)=f(an) 接下面问题补充
数列{bn}满足bn=绝对值(an-√3),sn=b1+b2+……+bn
(1)用数学归纳法证明bn
数列{bn}满足bn=绝对值(an-√3),sn=b1+b2+……+bn
(1)用数学归纳法证明bn
![已知函数f(x)=(x+3)/(x+1) (x不等于-1),设数列{an}满足a1=1,an+1(n+1是下标)=f(a](/uploads/image/z/16900439-23-9.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3D%28x%2B3%29%2F%28x%2B1%29+%28x%E4%B8%8D%E7%AD%89%E4%BA%8E-1%29%2C%E8%AE%BE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E6%BB%A1%E8%B6%B3a1%EF%BC%9D1%2Can%2B1%28n%2B1%E6%98%AF%E4%B8%8B%E6%A0%87%29%3Df%28a)
楼上有些复杂.
an-√3 = [a(n-1) +3]/ [a(n-1)+1] - 3 = (1-√3)* [a(n-1) -√3] / [a(n-1)+1] (1)
f(x)>1,所以 an=f (a(n-1) )> a(n-1) ,所以 an> a(n-1)> a(n-2) >……>a2>a1=1
所以1 / [a(n-1)+1]
an-√3 = [a(n-1) +3]/ [a(n-1)+1] - 3 = (1-√3)* [a(n-1) -√3] / [a(n-1)+1] (1)
f(x)>1,所以 an=f (a(n-1) )> a(n-1) ,所以 an> a(n-1)> a(n-2) >……>a2>a1=1
所以1 / [a(n-1)+1]
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