Question

已知数列{an}的前n项和Sn=-an-(1/2)^n-1+2(n为正整数)。①证明:an+1=(1/2)an+(1/2)^n+1,并求数列{an}的...

已知数列{an}的前n项和Sn=-an-(1/2)^n-1+2(n为正整数)。①证明:an+1=(1/2)an+(1/2)^n+1,并求数列{an}的通项:②若cn/n+1=an/n,Tn=c1+c2+…+cn,求Tn.

Answer 1

Sn=-an-1/2^n-1+2(n>=2).......1
Sn-1=-a(n-1)-1/2^(n-2)+2.......2
1-2得：an=an-1-an-1/2(n-2)
an=a(n-1)/2-1/2(n-1)
上式左右同乘以2^n得
2^nan=2^(n-1)a(n-1)-2
即bn=b(n-1)-2
即bn为等差数列。

Answer 2

①Sn=-an-(1/2)^(n-1)+2
S(n+1)=-a(n+1)-(1/2)^n+2
a(n+1)=S(n+1)-Sn
=-a(n+1)-(1/2)^n+2-[-an-(1/2)^(n-1)+2]
=an-a(n+1)+(1/2)^n
2a(n+1)=an+(1/2)^n
∴a(n+1)=1/2an+(1/2)^(n+1)
a1=S1=-a1-(1/2)^0+2
a1=1/2
an=1/2a(n-1)+(1/2)^n
a2=1/2*1/2+(1/2)^2=2/4
a3=1/2*1/2+(1/2)^3=3/8
a4=1/2*3/8+(1/2)^4=4/16
.....
an=n/2^n
②cn=an*(n+1)/n=(n+1)/2^n
Tn=2/2+3/2^2+4/2^3+...+(n+1)/2^n
Tn/2=2/2^2+3/2^3+4/2^4+...+(n+1)/2^(n+1)
Tn-Tn/2=2/2+1/2^2+1/2^3+...+1/2^n-(n+1)/2^(n+1)
Tn=2+(1/2+1/4+1/8+...+1/2^n)-(n+1)/2^(n+1)
=2+1/2*(1-1/2^n)/(1-1/2)-(n+1)/2^(n+1)
=3-1/2^n-(n+1)/2^(n+1)
=3-(n+3)/2^(n+1)