Question

已知数列an前n项和为sn=-an-(1/2)n-1+2①证明:an+1=(1/2)an+(1/2)^n+1,并求数列{an}的

通项公式 2，若cn/n+1=an/n 且tn=c1+c2+c3...+cn 求tn

Answer 1

(1)
n≥2时，an=Sn-S(n-1)=-an-(1/2)^(n-1) +2-[-a(n-1)-(1/2)^(n-2)+2]

2an=a(n-1)+(1/2)^(n-1)
an=(1/2)a(n-1)+(1/2)ⁿ
a(n+1)=(1/2)an+(1/2)^(n+1)，等式成立。
等式两边同乘以2^(n+1)
2^(n+1)·a(n+1)=2ⁿ·an +1
2^(n+1)·a(n+1)-2ⁿ·an=1，为定值
n=1时，a1=S1=-a1-(1/2)^0 +2
2a1=1
a1=1/2
2×a1=2×(1/2)=1，数列{2ⁿ·an}是以1为首项，1为公差的等差数列
2ⁿ·an=1+1×(n-1)=n
an=n/2ⁿ
数列{an}的通项公式为an=n/2ⁿ
(2)
cn/(n+1)=an/n
cn=[(n+1)/n]an=[(n+1)/n](n/2ⁿ)=(n+1)/2ⁿ
Tn=c1+c2+...+cn=2/2+3/2²+4/2³+...+(n+1)/2ⁿ
Tn/2=2/2²+3/2³+...+n/2ⁿ+(n+1)/2^(n+1)
Tn-Tn/2=Tn/2=2/2+1/2²+1/2³+...+1/2ⁿ- (n+1)/2^(n+1)
Tn=2+1/2+1/2²+...+1/2^(n-1) -(n+1)/2ⁿ
=[1+1/2+...+1/2^(n-1)] -(n+1)/2ⁿ +1
=1×(1-1/2ⁿ)/(1-1/2) -(n+1)/2ⁿ +1
=3- (n+3)/2ⁿ

Answer 2

Sn=-an-1/2^n-1+2
Sn+1=-an+1-1/2^n+2
Sn+1-Sn=an+1=-an+1+an+1/2^n
2an+1=an+1/2^n，即
an+1=(1/2)an+(1/2)^（n+1）,
2^(n+1)an+1=2^nan+1
bn+1=bn-2 数列{bn}是等差数列
an=n/2^n
(2)
Cn=(n+1)/2^n
Tn= 2/2+3/4+4/8.............n/2^n-1+(n+1)/2^n
2Tn=2+3/2+4/4.....................+(n+1)/2^n-1
2Tn-Tn=2+1/2+1/4.....................+1/2^n-1-(n+1)/2^n
=2+(1-1/2^n-1)-(n+1)/2^n
=3-1/2^n-1-(n+1)/2^n