Question

已知数列{an}满足a1=1,a(n+1)=an/(an+2)，(n属于N*)，若

若b(n+1)=(n-x)(1/an +1)，b1=-x，且数列{bn}是递增数列，则实数x的取值范围为( )Ax<2，Bx>3，Cx>2，Dx<3

Answer 1

a(n+1)=an/(an+2)

1/a(n+1)=(an+2)/an=1+ 2/an
1/a(n+1) +1=2 +2/an=2(1+1/an)
[1/a(n+1) +1]/(1+1/an)=2，为定值
1/a1+1=1+1=2，数列{1/an +1}是以2为首项，2为公比的等比数列
1/an +1=2ⁿ
b(n+1)=(n-x)(1/an+1)=(n-x)2ⁿ
n≥2时，b(n+1)>bn
b(n+1)-bn>0
b(n+1)-bn=(n-x)2ⁿ- (n-1-x)2^(n-1)
=(n-x+1)2^(n-1)>0
2^(n-1)恒>0，因此只有n-x+1>0
x<n+1
随n增大，n+1单调递增，要不等式对任意n≥2恒成立，只要n=2时成立
x<3
n=1时，b2=2(1-x)=2-2x
数列递增，a2>a1
2-2x>-x
x<2
综上，得x<2，选A
具体过程写了这么多，作为选择题，你实际做的时候不要写这么细，演草纸上大概步骤列一下就可以了。

Answer 2

a(n+1)=an/(an+3)
a(n+1)+2 =an/(an+3)+2
= 3(an+2)/(an+3)
1/[a(n+1)+2] = (an+3)[3(an+2)]
= 1/3 + (1/3)[ 1/(an+2)]
1/[a(n+1)+2] -1/2 = (1/3) { [ 1/(an+2)] - 1/2 }
=>{ [ 1/(an+2)] - 1/2 }是等比数列, q=1/3
{ [ 1/(an+2)] - 1/2 }= (1/3)^(n-1).{ [ 1/(a1+2)] - 1/2 }
=-(1/2).(1/3)^n
1/(an+2) =(1/2)( 1- (1/3)^n)
an +2 = 2/( 1- (1/3)^n)
an = -2 +2/( 1- (1/3)^n)
= -2 + 2.3^n/(3^n -1)
= 2/(3^n -1)

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)

bn=(3^n-1)n/(2^n*an)
=n/2^(n+1)
= (1/2)[ n. (1/2)^n ]
Tn =b1+b2+...+bn
=(1/2)S
= (1- (1/2)^n ) - n.(1/2)^(n+1)
= 1- (n+2)(1/2)^(n+1)