Sn=n^2an (N>=1)
Sn-1 =(n-1)^2a(n-1)
則Sn-Sn-1 = n^2an - (n-1)^2a(n-1)
an = n^2an - (n-1)^2a(n-1)
(n+1)(n-1)an=(n-1)^2a(n-1)
當(dāng)n=1時(shí),S1=1^2a1成立
當(dāng)n>1時(shí)：an/a(n-1)=(n-1)/(n+1)
列舉如下:
a2/a1 = 1/3
a3/a2 = 2/4
a4/a3 = 3/5
a5/a4 = 4/6
a6/a5 = 5/7
...
a(n-1)/a(n-2)=(n-2)/n
an/a(n-1)=(n-1)/(n+1)
左右全部相乘得
an/a1 = 2/n(n+1)
a1=1/2
則an=1/n(n+1)