Question

已知数列{an}满足：a1=2/3,an+1=1/(2-an).

Answer 1

1,

构造等比数列此烂丛
a(n+1)+3=2(an+3)
(即设bn=an+3)
上式就成了b(n+1)=2bn
（历袜首项b1=a1+3，公比=2等比数列）
bn=b1*q^(n-1)

an=5×2^(n-1)-3.

2,

{nan}=5n2^(n-1)-3n

sn=5×1×2^0-3×1+5×2×2^1-3×2+……+5n2^(n-1)-3n

sn=5×1×2^0+5×2×2^1+……+5n2^(n-1)-(3+3×2+3×3+……+3n)

sn=5×1×2^0+5×2×2^1+……+5n2^(n-1)-(3n^2+3n)/2

2sn=5×2×2^0+5×4×2^1+……+5n2^n-(3n^2+3n)

错森樱位相减得

-sn=5×2^0+5×2^1+5×2^2+……+5×2^(n-1)-5n2^n+(3n^2+3n)/2

-sn=5×(2^0+2^1+2^2+……+2^(n-1))-5n2^n+(3n^2+3n)/2

-sn=5×2^n-1-5n2^n+(3n^2+3n)/2

sn=5×2^n(n-1)-3n^2/2-3n/2

Answer 2

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。耐腊
1/(a1-1)=1/(2/3
-1)=-3
数列{1/(an
-1)}是以-3为首项，-1为公并陆差的
等差数列
。
(2)
1/(an
-1)=-3+(-1)(n-1)=-n-2
an
-1=1/(-n-2)=-1/绝亩顷(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