∵s[n]=n^2a[n]
∴s[n+1]=(n+1)^2a[n+1]
将上述两式相减,得：
a[n+1]=(n+1)^2a[n+1]-n^2a[n]
(n^2+2n)a[n+1]=n^2a[n]
即：a[n+1]/a[n]=n/(n+2)
于是：
a[n+1]/a[n]=n/(n+2)
a[n]/a[n-1]=(n-1)/(n+1)
a[n-1]/a[n-2]=(n-2)/n
a[n-2]/a[n-3]=(n-3)/(n-1)
.
a[5]/a[4]=4/6
a[4]/a[3]=3/5
a[3]/a[2]=2/4
a[2]/a[1]=1/3
将上述各项左右各自累乘,得：
a[n+1]/a[1]=(1*2)/[(n+1)(n+2)]
∵a[1]=1/2
∴a[n+1]=1/[(n+1)(n+2)]
∴通项a[n]=1/[n(n+1)]