早教吧作业答案频道 -->数学-->
已知数列{an}的前n项和为Sn,且a1=1/2,a(n+1)=(n+1)an/2n,(1)求{an}的通项公式;(2)设bn=n(2-Sn),n属于N*,若集合M={n|bn大于等于μ,n属于N*}恰有4个元素,求实数μ的取值范围.
题目详情
已知数列{an}的前n项和为Sn,且a1=1/2,a(n+1)=(n+1)an/2n,(1)求{an}的通项公式;(2)设bn=n(2-Sn),n属于N*,若集合M={n|bn大于等于μ,n属于N*}恰有4个元素,求实数μ的取值范围.
▼优质解答
答案和解析
(1)
a(n+1)=(n+1)an/(2n)
a(n+1)/(n+1) = (1/2) (an/n)
{an/n} 是等比数列,q=1/2
an/n = (1/2)^(n-1) .( a1/1)
= (1/2)^n
an = n.(1/2)^n
(2)
let
S = 1.(1/2)^1+2(1/2)^2+.+n.(1/2)^n (1)
(1/2)S = 1.(1/2)^2+2(1/2)^3+.+n.(1/2)^(n+1) (2)
(1) -(2)
(1/2)S = (1/2 + 1/2^2+...+1/2^n)-n(1/2)^(n+1)
= (1-1/2^n) - n(1/2)^(n+1)
S = 2 - (n+2)(1/2)^n
Sn =a1+a2+...+an
= S
= 2 - (n+2)(1/2)^n
bn = n(2-Sn)
= n(n+2)(1/2)^n
let
f(x) = x(x+2) (1/2)^x
f'(x) =( -x(x+2)ln2 + (2x+2) ) (1/2)^x =0
-x(x+2)ln2 + (2x+2)=0
(ln2)x^2 -(2-2ln2)x - 2 =0
x = 1.31
b1= 3(1/2)^1 = 3/2
b2 = 8(1/2)^2 = 2
max bn= b2 = 2
b3 = 15(1/8) = 15/8
b4 = 24(1/16) = 3/2
b5 = 35/32
M={n|bn>μ,n属于N*}恰有4个元素
35/32
a(n+1)=(n+1)an/(2n)
a(n+1)/(n+1) = (1/2) (an/n)
{an/n} 是等比数列,q=1/2
an/n = (1/2)^(n-1) .( a1/1)
= (1/2)^n
an = n.(1/2)^n
(2)
let
S = 1.(1/2)^1+2(1/2)^2+.+n.(1/2)^n (1)
(1/2)S = 1.(1/2)^2+2(1/2)^3+.+n.(1/2)^(n+1) (2)
(1) -(2)
(1/2)S = (1/2 + 1/2^2+...+1/2^n)-n(1/2)^(n+1)
= (1-1/2^n) - n(1/2)^(n+1)
S = 2 - (n+2)(1/2)^n
Sn =a1+a2+...+an
= S
= 2 - (n+2)(1/2)^n
bn = n(2-Sn)
= n(n+2)(1/2)^n
let
f(x) = x(x+2) (1/2)^x
f'(x) =( -x(x+2)ln2 + (2x+2) ) (1/2)^x =0
-x(x+2)ln2 + (2x+2)=0
(ln2)x^2 -(2-2ln2)x - 2 =0
x = 1.31
b1= 3(1/2)^1 = 3/2
b2 = 8(1/2)^2 = 2
max bn= b2 = 2
b3 = 15(1/8) = 15/8
b4 = 24(1/16) = 3/2
b5 = 35/32
M={n|bn>μ,n属于N*}恰有4个元素
35/32
看了 已知数列{an}的前n项和为...的网友还看了以下:
设A是n阶实对称矩阵,以下命题与"A是正定矩阵"不等价的是A.A的主对角线元素全部大于0B.存在实 2020-05-14 …
已知集合A={m,n/m,1},集合B={m²,m+n,0},若A=B,则()已知集合A={m,n 2020-05-16 …
交点圆锥曲线问题,哥哥姐姐们帮帮忙,集合M={(x,y)|x^2+(y-a)^2=1,x∈R,y∈ 2020-05-23 …
下列加点字注音有错误的一项是A.地壳(qiào)粮囤(dùn)切合实际(qiè)丢车保帅(jū)B 2020-06-16 …
设A={a1,a2,…,an}⊆M(n∈N*,n≥2),若a1+a2+…+an=a1a2…an,则 2020-07-21 …
若n为合数,n|x^2-1,则gcd(x+1,n)|ngcd(x-1,n)|n且gcd(x+1,n 2020-07-30 …
已知集合M={x|3x∧2-mx+2=0},N={x|mx=n},若集合M,N均只有一个元素,且M 2020-07-30 …
(1)m是什么实数时,方程mx2-(1-m)x+m=0没有实数根?...(x2是x的平方,不是mx 2020-08-01 …
1{x|x=2n,n∈Z}{x|x=4n+2,n∈Z}2设集合M={a,b},若集合N={x|x是M 2020-11-01 …
下列推断中,符合实际的是()A.第n周期的最后一种金属元素位于第n主族(n>1)B.第n周期有(8- 2020-12-07 …