已知等差数列{an}满足 a1+a(2n-1)=2n设Sn是数列{1/an}的前n项和 谁知道这个题从哪来的?
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已知等差数列{an}满足 a1+a(2n-1)=2n设Sn是数列{1/an}的前n项和 谁知道这个题从哪来的?
已知等差数列{an}满足 a1+a(2n-1)=2n设Sn是数列{1/an}的前n项和,记f(n)=S2n-Sn(n属于自然数)
(1)求an
(2)求f(n)的最小值
(3)如果函数g(x)=log2x-12f(n)(x属于[a,b])对任意自然数n,其函数值都恒小于零,那么a,b应满足什么条件
已知等差数列{an}满足 a1+a(2n-1)=2n设Sn是数列{1/an}的前n项和,记f(n)=S2n-Sn(n属于自然数)
(1)求an
(2)求f(n)的最小值
(3)如果函数g(x)=log2x-12f(n)(x属于[a,b])对任意自然数n,其函数值都恒小于零,那么a,b应满足什么条件
1.
设an=a1+(n-1)d
2n=a1+a(2n-1)=a1+a1+(2n-2)d
n=a1+(n-1)d=an
an=n;
2.
1/an=1/n
sn=1/1+1/2++1/3+……+1/n
s(2n)=1/1+1/2++1/3+……+1/n+1/(n+1)+……+1/(2n-1)+1/(2n)
f(n)=S2n-Sn
=1/(n+1)+……+1/(2n-1)+1/(2n)
≥1/(n+1)+……+1/(n+1)+1/(n+1)……(n项)
=n/(n+1)
=1-1/(n+1)
≥1-1/2=1/2
f(n)的最小值为1/2;
3.
g(x)=log2x-12[1/(n+1)+……+1/(2n-1)+1/(2n)]
=log2x-12f(n)
≤log2x-12*(1/2)
=log2x-6
<0
log2x<6
1<x<2^6=64
所以1<a<b<64.
设an=a1+(n-1)d
2n=a1+a(2n-1)=a1+a1+(2n-2)d
n=a1+(n-1)d=an
an=n;
2.
1/an=1/n
sn=1/1+1/2++1/3+……+1/n
s(2n)=1/1+1/2++1/3+……+1/n+1/(n+1)+……+1/(2n-1)+1/(2n)
f(n)=S2n-Sn
=1/(n+1)+……+1/(2n-1)+1/(2n)
≥1/(n+1)+……+1/(n+1)+1/(n+1)……(n项)
=n/(n+1)
=1-1/(n+1)
≥1-1/2=1/2
f(n)的最小值为1/2;
3.
g(x)=log2x-12[1/(n+1)+……+1/(2n-1)+1/(2n)]
=log2x-12f(n)
≤log2x-12*(1/2)
=log2x-6
<0
log2x<6
1<x<2^6=64
所以1<a<b<64.
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