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}...

发布网友 发布时间：2024-03-04 00:18

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热心网友 时间：2024-03-27 13:16

解：
(1)
a(n+1)=(n+1)an/(2n)
a(n+1)/(n+1)=(1/2)(an/n)
[a(n+1)/(n+1)]/(an/n)=1/2，为定值
a1/1=(1/2)/1=1/2，数列{an/n}是以1/2为首项，1/2为公比的等比数列
an/n=(1/2)(1/2)^(n-1)=1/2ⁿ
an=n/2ⁿ
数列{an}的通项公式为an=n/2ⁿ
(2)
Sn=a1+a2+a3+...+an=1/2+2/2²+3/2³+...+n/2ⁿ
Sn /2=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
Sn -Sn/2=Sn /2=1/2+1/2²+...+1/2ⁿ -n/2^(n+1)
=(1/2)(1-1/2ⁿ)/(1-1/2) -n/2^(n+1)
=1- (n+2)/2^(n+1)
Sn=2- (n+2)/2ⁿ
bn=n(2-Sn)=n[2-2+(n+2)/2ⁿ]=n(n+2)/2ⁿ
b1=1×3/2=3/2 b2=2×4/4=2 b3=3×5/8=15/8<b1
n≥2时，
b(n+1)/bn=[(n+1)(n+3)/2^(n+1)]/[n(n+2)/2ⁿ]
=(n+1)(n+3)/[2n(n+2)]
=(n²+4n+3)/(2n²+4n)
=(1/2)(2n²+4n+4n+6)/(2n²+4n)
=(1/2)[1 +(2n+3)/(n²+2n)]
(2n+3)/(n²+2n) -1
=(2n+3-n²-2n)/(n²+2n)
=(3-n²)/(n²+2n)
n≥2 n²≥4 3-n²<0 b(n+1)<bn，即数列从第2项开始，单调递减。
bn≥μ n(n+2)/2ⁿ≥μ
集合M恰有4个元素，又b3<b1，因此集合M={b1，b2，b3，b4}
b4≥μ b5<μ
5×(5+2)/2^5<μ≤4×(4+2)/2⁴
35/32<μ≤3/2
μ的取值范围为(35/32，3/2]

热心网友 时间：2024-03-27 13:09

(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<μ< 3/2