Question

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

a1=1，a2=1/2，a（n+2）=[a(n+1)]^2/an+a(n+1)

求证：数列{an/a（n+1)} 是等差数列：

求数列an的通项公式

Answer 1

求an？

an+2=(an+1)^2/（an+an+1）
两边取倒数
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

那个比较长的式子打括号写的清楚点

Answer 3

问什么？