已知数列满足a1=1,2a(n+1)an+a(n+1)-an=0,(1)求证:{1/an}是等差数列.(2)若a1a2+a2a3+…+ana(n+1)>16/33求n的取值范围
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![已知数列满足a1=1,2a(n+1)an+a(n+1)-an=0,(1)求证:{1/an}是等差数列.(2)若a1a2+a2a3+…+ana(n+1)>16/33求n的取值范围](/uploads/image/z/8907715-19-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%E6%BB%A1%E8%B6%B3a1%3D1%2C2a%28n%2B1%29an%2Ba%28n%2B1%29-an%3D0%2C%281%29%E6%B1%82%E8%AF%81%EF%BC%9A%7B1%2Fan%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97.%282%29%E8%8B%A5a1a2%2Ba2a3%2B%E2%80%A6%2Bana%28n%2B1%29%3E16%2F33%E6%B1%82n%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
已知数列满足a1=1,2a(n+1)an+a(n+1)-an=0,(1)求证:{1/an}是等差数列.(2)若a1a2+a2a3+…+ana(n+1)>16/33求n的取值范围
已知数列满足a1=1,2a(n+1)an+a(n+1)-an=0,(1)求证:{1/an}是等差数列.(2)若a1a2+a2a3+…+ana(n+1)>16/33
求n的取值范围
已知数列满足a1=1,2a(n+1)an+a(n+1)-an=0,(1)求证:{1/an}是等差数列.(2)若a1a2+a2a3+…+ana(n+1)>16/33求n的取值范围
(1)
2a(n+1)an+a(n+1)-an=0
两边同时除以a(n+1)an
2+1/an-1/a(n+1)=0
即1/a(n+1)-1/an=2
所以数列{1/an}为等差数列,公差d=2
首项1/a1=1
∴1/an=1/a1+(n-1)d=1+2(n-1)=2n-1
∴an=1/(2n-1)
(2)
an*a(n+1)=1/2[an-a(n+1)]=1/2[1/(2n-1)-1/(2n+1)]
∴a1a2+a2a3+…+ana(n+1)
=1/2{(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+[1/(2n-1)-1/(2n+1)]}
=1/2[1-1/(2n+1)]
∵a1a2+a2a3+…+ana(n+1)>16/33
∴1/2[1-1/(2n+1)]>16/33
∴1-1/(2n+1)>32/33
∴1/(2n+1)33
n>16
(1)an-a(n+1)=2a(n+1)an
1/a(n+1)-1/an=2
{1/an}是以1为首项,以2为公差的等差数列
(2)由{1/an}是以1为首项,以2为公差的等差数列可知
1/an=2n-1
an=1/(2n-1)
ana(n+1)=1/[(2n-1)1/(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
a1a2+a...
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(1)an-a(n+1)=2a(n+1)an
1/a(n+1)-1/an=2
{1/an}是以1为首项,以2为公差的等差数列
(2)由{1/an}是以1为首项,以2为公差的等差数列可知
1/an=2n-1
an=1/(2n-1)
ana(n+1)=1/[(2n-1)1/(2n+1)]=1/2[1/(2n-1)-1/(2n+1)]
a1a2+a2a3+…+ana(n+1)=1/2[1-1/3+1/3-1/5+....+1/(2n-1)-1/(2n+1)]
=1/2[1-1/(2n+1)]
=n/2n+1
n/2n+1>16/33
n>16
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