Question

已知数列{an}满足:a1=1,a2=1/2,an+2=(an+1)^2/（an+an+1）

已知数列{an}满足：a1=1,a2=1/2,an+2=(an+1)^2/（an+an+1）（a后面的均为下标）
1.求证数列{an/an+1)是等差数列
2.求数列{an}的通项公式。
过程麻烦详细点。

Answer 1

an+2=(an+1)^2/（an+an+1）
2 边取倒数
1/a(n+2)=[an+a(n+1)]/[a(n+1)*a(n+1)]
a(n+1)/a(n+2)=[an+a(n+1)]/a(n+1)
= an/a(n+1) + 1
设bn=an/a(n+1) 则 b(n+1)=a(n+1)/a(n+2)
b(n+1)=bn+1 b(n+1)-bn=1
==> bn 即{an/a(n+1)} 为等差数列 ,首项为 b1=a1/a2=2 d=1
bn = an / a(n+1) = b1 + (n-1) d = 2 +(n-1) = n+1
an/a(n+1) = n+1
a(n-1)/an = n
a(n-2)/a(n-1)= n-1
...
a2/a3 = 3
a1/a2 = 2
两边相乘
a1/a(n+1) = 2*3*4*5...*(n+1) =(n+1)!
a(n+1)=a1/(n+1)!=1/(n+1)!
==> an=1/[n!]

Answer 2

你可以去查,只要你查的到