Question

数列 {an}满足a1 =1/2 ,a1+a2+a3+......an=n^2an 则an=?

Answer 1

Sn=n^2an,所以Sn-1=(n-1)^2an-1,这两式相减得an=n^2an-(n-1)^an-1,所以（n^2-1)an=(n-1)^2an-1
两边同时约去(n-1),得（n+1）an=(n-1)an-1,所以an/an-1=(n-1)/(n+1),剩下的步骤就是数列中的累乘法的运用了，a2/a1=1/3,a3/a2=2/4,a4/a3=3/5……，an-2/an-3=(n-3)/(n-1),an-1/an-2=(n-2)/n,an/an-1=(n-1)/(n+1),上述（n-1）个式子相乘得an/a1=(1*2)/[n(n+1)]，把a1代入即可。不懂可问。

Answer 2

a1+a2+a3+...+an=n^2 an
a1+a2+a3+...+an-1=(n-1)^2 an-1
两式相减，得
an=n^2 an-(n-1)^2 an-1，
化简得 an/an-1=(n-1)^2/(n^2-1)=(n-1)/(n+1)
故an=(n-1)/(n+1) an-1
=(n-1)/(n+1) (n-2)/n an-2
=(n-1)/(n+1) (n-2)/n (n-3)/(n-1) an-3
= ...
=(n-1)/(n+1) (n-2)/n (n-3)/(n-1) ... 2/4 1/3 a1
=2/(n+1)n a1
=1/(n+1)n