Question

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)

Answer 1

(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
maxbn=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