Question

已知数列{an}中,a1=负1,An+1×An=An+1－An,则数列通项公式an=_____…求

已知数列{an}中,a1=负1,An+1×An=An+1－An,则数列通项公式an=_____…求详细解析！

Answer 1

A(n+1)×An=A(n+1)-An
两边同除A(n+1)×An
1=1/An-1/A(n+1)
1/A(n+1)-1/An=-1
数列{1/An}是等差数列
A1=-1
1/A1=-1
1/An=-1+(n-1)×(-1)=-n
An=-1/n

追问

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Answer 2

,a1=－1,a(n+1)×an=a(n+1)－an,
a(n+1)－an=a(n+1)×an
两边同除以a(n+1)×an得
1/a(n+1)]-[1/an]=-1
{1/an}是等差数列
1/an＝1/a1－（n-1）=-1－（n-1）=-n
an= -1/n

Answer 3

a(n+1)*an=a(n+1)-an

a(n+1)*(an-1)=-an
1/a(n+1)=1/an-1
1/a1=-1
所以{1/a(n+1)}是以-1为首项，-1为公差的等差数列
1/an=-n
an=-1/n