题目

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

提问时间：2021-01-03

答案

(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