Question

已知数列{an}的前n项的和为Sn 且向量a=（n,Sn）b=(4,n+3)共线

Answer 1

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)