Question

数列求解

设数列{An}的前n项和为Sn，若A1=4/3,An+1=3Sn,n属于正整数。（1）求数列{An}的通项公式；（2）设Bn=log2An+1，求数列{Bn}前n项和Tn.(3)令Cn=1/Tn,数列{Cn}的前n项和为Un,试求最小的集合［a,b),使Un属于[a,b).

Answer 1

a(n+1) = 3s(n) = s(n+1) - s(n),
s(n+1) = 4s(n),
{s(n)}是首项为s(1)=a(1)=4/3,公比为4的等比数列。
s(n) = (4/3)4^(n-1) = 4^n/3,

a(n+1) = 3s(n) = 4^n,

a(1) = 4/3,
n>=2时，a(n) = 4^(n-1).

b(n) = log_{2}[a(n+1)] = log_{2}4^n = 2n,
t(n) = n(n+1).

c(n) = 1/t(n) = 1/n - 1/(n+1).
u(n) = c(1)+c(2)+...+c(n-1)+c(n)
= 1/1 - 1/2 + 1/2 - 1/3 + ... + 1/(n-1) - 1/n + 1/n - 1/(n+1)
= 1/1 - 1/(n+1)
= n/(n+1)
= 1/(1+1/n)

n>=1时，
1/n单调递减。。u(n) = 1/(1+1/n)单调递增。。
1/2 <= u(1) <= u(n) < 1.

a = 1/2,
b = 1.