Question

求解析过程，谢谢

求解析过程，谢谢在线等急

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Answer 1

三
(1)an=3Sn-2 , a1=S1=1
a(n-1)=3S(n-1)-2
an-a(n-1)=3[Sn-S(n-1)]
an-a(n-1)=3an
an/a(n-1)=-1/2
an=(-1/2)^(n-1)
(2)Sn=1+(-1/2)+(-1/2)^2+(-1/2)^3+...+(-1/2)^(n-1)=[(1-(-1/2)^n]/(1+1/2)
=2/3[(1-(-1/2)^n]
Sn=2/3[(1-(-1/2)^n]
四
(1)Sn+1/2*an=1 S1=a1 ,a1=2/3
S(n-1)+1/2*a(n-1)=1
[Sn-S(n-1)]+1/2*an-1/2*a(n-1)=0
an+1/2*an-1/2*a(n-1)=0
an/a(n-1)=1/3
an=(2/3)*(1/3)^(n-1)=2*(1/3)^n
an=2*(1/3)^n
(2)
S(n+1)=2[1/3+(1/3)^2+...+(1/3)^(n+1)]=2*[1/3(1-(1/3)^(n+1)]/(1-1/3)
=1-(1/3)^(n+1)
bn=log(1/3)[1-S(n+1)]=log(1/3)[(1/3)^(n+1)]=n+1
b(n+1)=n+2
1/[bnb(n+1)]=1/(n+1)(n+2)=[1/(n+1)]-[1/(n+2)]

Tn=[(1/2-1/3)+(1/3-1/4)+...+1/(n+1)]-[1/(n+2)]=1/2-1/(n+2)
Tn=n/[2(n+2)]