在数列{a1}中,已知a1=1,Sn=n2an求通项公式an

Question

在数列{a1}中,已知a1=1,Sn=n2an求通项公式an

Answer 1

a[1]=1, S[n]=n^2*a[n]
答案是 a[n]=6/((n+1)(n+2))
推导如下:
S[n]=n^2*a[n]
S[n-1]=(n-1)^2*a[n-1]
a[n]=S[n]-S[n-1]=n^2*a[n]-(n-1)^2*a[n-1]
(n^2-1)*a[n]=(n-1)^2*a[n-1]
(n^2-1)*a[n]=(n-1)^2*a[n-1]
a[n]/a[n-1]=(n-1)^2/(n^2-1)
a[n]/a[n-1]=(n-1)/(n+1)
a[n]=a[n-1](n-1)/(n+1)

a[1]=1 =2*3/(2*3)
a[2]=2/4 =2*3/(3*4)
a[3]=(2/4)*(3/5)=2*3/(4*5)
a[4]=(2/4)*(3/5)*(4/6)=(2*3)/(5*6)
...
a[n]=(2/4)*(3/5)*(4/6)*.*(n/(n+2))=2*3/((n+1)(n+2))
=6/((n+1)(n+2))