(n+1)na[n+1]=n(n-1)a[n]
n≧3
n(n-1)a[n]=(n-1)(n-2)a[n-1]
=(n-2)(n-3)a[n-2]
=・・・
=2*1*a[2]
=2(∵a[2]=1)

∴
a[n]=0 (n=1)
a[n]=2/n(n-1) (n≧2)