∵a(n+1)/an=（n+2）/n,a1=1,
∴an=(n+1)/(n-1)*a(n-1)
=(n+1)n/[(n-1)(n-2)]*a(n-2)
.
=(n+1)n(n-1).4*3/[(n-1)(n-2).2*1]*a1
=n(n+1)/2an=(n+1)/(n-1)*a(n-1)怎么来的把a(n+1)/an=（n+2）/n中的令n+1=m即a(m)/a(m-1)=（m+1）/(m-1)即an=(n+1)/(n-1) *a(n-1) [n相当与m]