已知数列an中,a1=-1,a(n+1)*an=a(n+1)-an,则数列通项公式an=?

a(n+1)*an=a(n+1)-an
两边除以a(n+1)*an得
1/a(n+1)-1/an=-1
令bn=1/an
则bn-b(n-1)=-1,b1=1/a1=-1
即bn是以-1为首项,-1为公差的等差数列
∴1/an=bn=-n,an=-1/n