a(n+1)=3an/(an+3)
a2=(3*1/2)/(1/2+3)=(3/2)/(7/2)=3/7
a3=(3*3/7)/(3/7+3)=(9/7)/(24/7)=9/24=3/8
a4=(3*3/8)/(3/8+3)=(9/8)/(27/8)=9/27=3/9
.
an=3/(n+5)
证明：（1）当n=1时,a1=3/6=1/2, 命题成立
（2）设n=k时,ak=3/(k+5)成立.
则n=k+1时有
a(k+1)=3ak/(ak+3)={9/[k+5]}/3[1/(k+5)+1]=[3/(k+5)]/[(k+6)/(k+5)]=3/(k+6)=3/[(k+1)+5]
所以a(k+1)=3/[(k+1)+5]
对于n∈N+都成立. a(n+1)=3an/(an+3)
a2=(3*1/2)/(1/2+3)=(3/2)/(7/2)=3/7
a3=(3*3/7)/(3/7+3)=(9/7)/(24/7)=9/24=3/8
a4=(3*3/8)/(3/8+3)=(9/8)/(27/8)=9/27=3/9
.
an=3/(n+5)
证明：（1）当n=1时,a1=3/6=1/2, 命题成立
（2）设n=k时,ak=3/(k+5)成立.
则n=k+1时有
a(k+1)=3ak/(ak+3)={9/[k+5]}/3[1/(k+5)+1]=[3/(k+5)]/[(k+6)/(k+5)]=3/(k+6)=3/[(k+1)+5]
所以a(k+1)=3/[(k+1)+5]
对于n∈N+都成立.