Question

已知数列{an}的前n项和为sn,且a1=1,an+1=二分之一乘sn(n=1.2.3....) (1)求数列{an}等等通项公式

(2)当bn=log二分之三(3an+1)时 求证:数列{bn*bn+1分之1}的前n项和,Tn=1+n分之n

Answer 1

∵a(n+1)=1/2*Sn,a1=1
∴a2=1/2*a1=1/2
a3=1/2*S2=1/2(a1+a2)=3/4
当n≥2时，an=1/2*S(n-1)
∴a(n+1)-an
=1/2*Sn-1/2*S(n-1)
=1/2*[Sn-S(n-1)]=1/2*an
∴a(n+1)=3/2*an
a(n+1)/an=3/2
∵a2=a1=1/2
∴{an}从第2项起为等比数列,公比为3/2
即n≥2时，an=a2*q^(n-2)=1/2*(3/2)^(n-2)
∴数列{an}等等通项公式 为分段形式
an={ 1, (n=1)
{ 1/2*(3/2)^(n-2)
(2)
∵ a(n+1)=1/2*(3/2)^(n-1)
∴3a(n+1)=(3/2)*(3/2)^(n-1)=(3/2)^n
∴bn=log(3/2)[3a(n+1)] =log(3/2)[(3/2)^n)=n
∴1/[(bnb(n+1)]=1/[n(n+1)]=1/n-1/(n+1)
∴Tn=(1-1/2)+(1/2-1/3)+(1/3-1/4)+.......+(1/n-1/(n+1))
=1-1/(n+1)
∴Tn=n/(n+1)

Answer 2

a1 = 1,
a2 = 1/2
a3 = 3/4
...
an = 1-1/2^(n-1)

追问

这是第几问啊,一共有两问呢,