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
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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
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=0](/uploads/image/z/16245417-57-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn+%E5%90%91%E9%87%8Fa%3D%28n%2CSn%29+b%3D%28n%2B3%2C-4%29+%E4%B8%94a%C2%B7b%3D0)
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)
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)
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