数列{an}满足a1=1,an=an-1(1+2an)(n-1是下标).1、求证:{1/an}是等差数列;2、若a1a2+a2a3+...+ana(n+1)>16/33
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数列{an}满足a1=1,an=an-1(1+2an)(n-1是下标).1、求证:{1/an}是等差数列;2、若a1a2+a2a3+...+ana(n+1)>16/33数列{an}满足a1=1,an=
数列{an}满足a1=1,an=an-1(1+2an)(n-1是下标).1、求证:{1/an}是等差数列;2、若a1a2+a2a3+...+ana(n+1)>16/33
数列{an}满足a1=1,an=an-1(1+2an)(n-1是下标).1、求证:{1/an}是等差数列;2、若a1a2+a2a3+...+ana(n+1)
>16/33
数列{an}满足a1=1,an=an-1(1+2an)(n-1是下标).1、求证:{1/an}是等差数列;2、若a1a2+a2a3+...+ana(n+1)>16/33
你的题目错了,下标是n+1,不是n-1
a(n)=a(n+1)(1+2an)
a(n)=a(n+1)+2a(n)a(n-1)
两边同时除以a(n)a(n-1)
1/a(n+1)=1/a(n)+2
1/a(n+1)-1/a(n)=2
(1)所以 {1/an}是等差数列
首项1/a1=1,公差为2
第二问题目也不全,可以利用裂项求和
(2)1/an=1+2(n-1)=2n-1
an=1/(2n-1)
an*a(n+1)=1/(2n-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
33n>16(2n+1)
n>16
所以 n的最小值为17
不会啊。
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