已知数列｛an}的前n项和为Sn 向量a=(n,Sn) b=(n+3,-4) 且a·b=01求证：数列｛an｝是等差数列2求数列｛1/nan｝的前n项和Tn

已知数列｛an}的前n项和为Sn 向量a=(n,Sn) b=(n+3,-4) 且a·b=01求证：数列｛an｝是等差数列2求数列｛1/nan｝的前n项和Tn
已知数列｛an}的前n项和为Sn 向量a=(n,Sn) b=(n+3,-4) 且a·b=0
1求证：数列｛an｝是等差数列
2求数列｛1/nan｝的前n项和Tn

已知数列｛an}的前n项和为Sn 向量a=(n,Sn) b=(n+3,-4) 且a·b=01求证：数列｛an｝是等差数列2求数列｛1/nan｝的前n项和Tn
a·b=0
n(n+3)+sn*(-4)=0
4sn=n(n+3)
sn=n(n+3)/4
s(n-1)=(n-1)(n+2)/4
an=sn-s(n-1)
=n(n+3)/4-(n-1)(n+2)/4
=(n^2+3n)/4-(n^2+n-2)/4
=(n^2+3n-n^2-n+2)/4
=(2n+2)/4
=(n+1)/2
a(n-1)=n/2
an-a(n-1)=(n+1)/2-n/2=1/2
所以an 是以1/2 为公差的等差数列
1/nan
=1/n[(n+1)/2]
=2/n(n+1)
=2*[1/n-1/(n+1)]
Tn
=2*(1-1/2)+2*(1/2-1/3)+.+2*[1/n-1/(n+1)]
=2*[1-1/2+1/2-1/3+.+1/n-1/(n+1)]
=2*[1-1/(n+1)]
=2n/(n+1)