已知Sn是数列an的前n项和,且a1=1/2,an/Sn=1/n2,则S100-a100=多少

a[n]/S[n]=1/n^2
S[n]=n^2*a[n]
S[n-1]+a[n]=n^2*a[n]
S[n-1]=(n^2-1)*a[n]
同时,有
S[n-1]=(n-1)^2*a[n-1]
所以
(n^2-1)*a[n]=(n-1)^2*a[n-1]
a[n]/a[n-1]=(n-1)^2/(n^2-1)=(n-1)/(n+1)
这样,就有
a[n]=1/(n(n+1))
S[n]=n^2*a[n]=n/(n+1)
所以 S[n]-a[n]=S[n-1]
所以 S[100]-a[100]
=S[99]
=99/100