已知数列(An)中,A1=1/3,AnAn-1=An-1-An(n>=2),数列Bn满足Bn=1/An,求数列Bn的通项

AnAn-1=An-1-An
AnAn-1+An=An-1
An=An-1/(A(n-1)+1) n>=2
A1=1/3
A2=A1/(A1+1) =1/3/(1/3+1)=1/4
A3=A2/(A2+1)=A1/(A1+1) / (A1/(A1+1)+1)
=A1/(A1+A1+1)=A1/(2A1+1)
=1/3/(2/3+1) =1/5
A4=A3/(A3+1)=A1/(2A1+1) / ( A1/(2A1+1)+1)
=A1/(A1+2A1+1)=A1/(3A1+1)
=1/3/(3/3+1)=1/6
.
An=A1/((n-1)A1+1)=1/3 / ((n-1)1/3 +1)
=1/(n-1+3)
=1/(n+2)
Bn=1/An=1/1/(n+2)=n+2

AnAn-1=An-1-An
(An-1-An)/AnAn-1=(1/An)-(1/An-1)=1
则Bn为B1=3 d=1 的等差数列
所以Bn=3+(n-1)*1＝n+2