第1个回答 2009-12-14
n≥2时:
a[n]=S[n]-S[n-1]=n^2a[n]-(n-1)^2a[n-1]
(n^2-1)a[n]=(n-1)^2a[n-1]
(n+1)a[n]=(n-1)a[n-1]
∴a[n]=(n-1)/(n+1)*a[n-1]
∴a[n]=(n-1)/(n+1)*a[n-1]
=(n-1)/(n+1)*(n-2)/n*a[n-2]
=…
=(n-1)/(n+1)*(n-2)/n*…*1/3*a[1]
=1/[n(n+1)]
考虑到a[1]=1/2
a[n]=1/[n(n+1)],n∈N.