Question

已知数列{an}满足a1=0,an+1=(n+2)/nan+1/n,an=——

Answer 1

解：
a(n+1)=[(n+2)/n]an+ 1/n
na(n+1)=(n+2)an +1
等式两边同除以n(n+1)(n+2)
a(n+1)/[(n+1)(n+2)]=an/[n(n+1)]+ 1/[n(n+1)(n+2)]
a(n+1)/[(n +1)(n+2)]=an/[n(n+1)]+½{1/[n(n+1)] -1/[(n+1)(n+2)]}
2a(n+1)/[(n+1)(n+2)]=2an/[n(n+1)]+ 1/[n(n+1)] -1/[(n+1)(n+2)]
[2a(n+1)+1]/[(n+1)(n+2)]=(2an+1)/[n(n+1)]
(2a₁+1)/[1·(1+1)]=(0+1)/2=½
数列{(2an+1)/[n(n+1)]}是各项均为½的常数数列
(2an+1)/[n(n+1)]=½
an=(n²+n-2)/4
n=1时，a₁=(1²+1-2)/4=0，同样满足表达式
数列{an}的通项公式为an=(n²+n-2)/4