向量a=(n,Sn)b=(4,n+3)共线
所以n(n+3)-4Sn=0
Sn=n(n+3)/4
a1=S1=1
n>=2:
an=Sn-S(n-1)=n(n+3)/4-(n-1)(n+2)/4=(n+1)/2
a1也符合,故an=(n+1)/2
1/nan=1/[n(n+1)/2]=2/(n(n+1))=2(1/n-1/(n+1))
前n项和Tn=2[1/1-1/2+1/2-1/3+...+1/n-1/(n+1))
=2[1-1/(n+1)]
=2n/(n+1)