Question

数学数列求解

Answer 1

f(x1)=2x1/(x1+2)=2/2013
x1+2=2013x1, x1=2/2012=1/1006
x(n+1)=f(xn)=2xn/(xn+2)
1/x(n+1)=(xn+2)/(2xn)=1/2+1/xn
即有1/x(n+1)-1/xn=1/2
故{1/xn}是以1/x1=1006为首项,公差d=1/2有等差数列,则有1/xn=1006+1/2(n-1)=1/2n+2011/2
故有xn=1/[1/2(n+2011)]=2/(n+2011)
(2)an=4/xn-4023=4/[2/(n+2011)]-4023=2n+4022-4023=2n-1
bn=(a(n+1)^2+an^2)/[2a(n+1)an]=[(2n+1)^2+(2n-1)^2]/[2(2n+1)(2n-1)]=(4n^2+1)/(2n+1)(2n-1)=1+2/(2n+1)(2n-1)=1+1/(2n-1)-1/(2n+1)
b1+b2+...+bn
=n+1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)
=n+1-1/(2n+1)
<n+1
得证!

Answer 2

(1)
f(x1)=2x1/(x1+2)=2/2013
4024x1=4
x1=1/1006
x(n+1)=f(xn)=2xn/(xn +2)
1/x(n+1)=(xn +2)/(2xn)=1/xn +1/2
1/x(n+1)-1/xn=1/2，为定值
1/x1=1/(1/1006)=1006，数列{1/xn}是以1006为首项，1/2为公差的等差数列
1/xn=1006+(1/2)(n-1)=(n+2011)/2
xn=2/(n+2011)
(2)
an=4/xn -4023=4/[2/(n+2011)] -4023=2n-1
bn=[a(n+1)²+an²]/[2a(n+1)an]
=[(2n+1)²+(2n-1)²]/[2(2n+1)(2n-1)]
=(4n²+1)/[(4n²-1)]
=(4n²-1+2)/(4n²-1)
=1+ 2/[(4n²-1)]
=1+ 2/[(2n+1)(2n-1)]
=1+ 1/(2n-1) -1/(2n+1)
b1+b2+...+bn=1+1/1-1/3+1+1/3-1/5+...+1+1/(2n-1)-1/(2n+1)
=n+[1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=n+1 -1/(2n+1)
1/(2n+1)>0 n+1 -1/(2n+1)<n+1
b1+b2+...+bn<n+1

Answer 3