有关高一数列的

题目

有关高一数列的
1、在数列{An}中A1=1,A(n+1)=(1+1/n)An+(n+1)/2^n.（1）求该数列的通项公式；（2）求该数列的前n项和Sn.
2、数列{An}中Sn=-An-(1/2)^(n-1)+2.（1）求该数列的通项公式；（2）若Bn=(n+1)An/n,Tn=B1+B2+……+Bn,求证Tn＜3
注：A(n+1)即数列{An}的第n+1项,B1、B2……即数列{Bn}的第1、2……项

答案

1.由A1=1 A(n+1)=(1+1/n)An+(n+1)/2^n得
A(n+1)/(n+1)-An/n=1/2^n
设Bn=An/n B(n+1)=A(n+1)/(n+1) B1=A1=1
则B(n+1)-Bn=1/2^n
Bn-B(n-1)=1/2^(n-1)
.
B2-B1=1/2
叠加Bn-B1=1/2+1/2^2+...+1/2^(n-1)=(1/2)[1-1/2^(n-1)]/(1-1/2)=1-1/2^(n-1)
Bn=2-1/2^(n-1)
An=nBn=2n-n/2^(n-1)
设Sn=2∑n-∑n/2^(n-1)=n(n+1)-Tn
Tn=1+2/2^1+...+n/2^(n-1)
(1/2)Tn=1/2^1+2/2^2+...+n/2^n
错位相减(1/2)Tn=1+1/2^1+1/2^2+...+1/2^(n-1)-(n+1)/2^n=(1-1/2^n)/(1-1/2)-n/2^n
=2-2/2^n-n/2^n=2-(n+2)/2^n
∴Tn=4-(n+2)/2^(n-1)
∴Sn=n^2+n+4-(n+2)/2^(n-1)
2.Sn=-An-1/2^(n-1)+2 S1=a1=-a1-1+2 a1=1/2
S(n-1)=-A(n-1)-1/2^(n-2)+2
∴An=Sn-S(n-1)=A(n-1)-An+1/2^(n-1)
∴An*2^n-A(n-1)*2^(n-1)=1
即{An*2^n}是公差为1的等差数列
An*2^n=2A1+n-1=n
∴(1) An=n/2^n
(2) Bn=(n+1)An/n=(n+1)/2^n
Tn=2/2^1+3/2^2+...+(n+1)/2^n
(1/2)Tn=2/2^2+3/2^3+...+(n+1)/2^(n+1)
错位相减(1/2)Tn=2/2^1+1/2^2+...+1/2^n-(n+1)/2^(n+1)=1/2+(1/2)*(1-1/2^n)/(1-1/2)-(n+1)/2^(n+1)
=1/2+1-1/2^n-(n+1)/2^(n+1)
∴Tn=3-(n+1)/2^n-2/2^n=3-(n+3)/2^n
∴Tn