Question

已知数列{an}满足：a1=2/3,an+1=1/(2-an).

已知数列{an}满足：a1=2/3,an+1=1/(2-an).(1)求证：数列{1/(an-1)}是等差数列；(2)令bn=[(n+2)/2n]xan,Tn=b1+b2+......+bn,试求Tn.

Answer 1

解：
(1)
a(n+1)=1/(2-an)
a(n+1) -1=(1-2+an)/(2-an)=(an -1)/(2-an)
1/[a(n+1) -1]=(2-an)/(an -1)=(1-an +1)/(an -1)=1/(an -1) -1
1/[a(n+1) -1]-1/(an -1)=-1，为定值。
1/(a1-1)=1/(2/3 -1)=-3
数列{1/(an -1)}是以-3为首项，-1为公差的等差数列。

(2)
1/(an -1)=-3+(-1)(n-1)=-n-2
an -1=1/(-n-2)=-1/(n+2)
an=1- 1/(n+2)=(n+1)/(n+2)
bn=[(n+2)/2ⁿ]×an=[(n+2)/2ⁿ](n+1)/(n+2)]=(n+1)/2ⁿ=n/2ⁿ +1/2ⁿ
Tn=(1/2+2/2²+...+n/2ⁿ) +(1/2+1/2²+...+1/2ⁿ)
令Cn=1/2+2/2²+...+n/2ⁿ
则Cn/2=1/2²+2/2³+...+(n-1)/2ⁿ+n/2^(n+1)
Cn-Cn/2=Cn/2=1/2+1/2²+...+1/2ⁿ-n/2^(n+1)
Cn=1+1/2+1/2²+...+1/2^(n-1) -n/2ⁿ
Tn=Cn+(1/2+1/2²+...+1/2ⁿ)
=1+1/2+1/2²+...+1/2^(n-1) -n/2ⁿ+(1/2+1/2²+...+1/2ⁿ)
=1×(1-1/2ⁿ)/(1-1/2) -n/2ⁿ +(1/2)(1-1/2ⁿ)/(1-1/2)
=3 -(n+3)/2ⁿ

Answer 2

120+564987987651354+1232498=1324987206374951