Question

高中数学，数列题目，在线等

Answer 1

分析：

（Ⅰ）根据条件，将递推关系式转化为项与项的关系，由等比数列定义得到t的值；
（Ⅱ）由（Ⅰ）得到新数列通项，再通过错位相减法进行求和，证出不等关系成立．

 

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Answer 2

Answer 3

an+1 = 2Sn + 1 (1)
an = 2sn-1+1 (2) n>=2
=>an+1-an = 2an n>=2
=>an+1 = 3an n>=2
等比数列，a2 = 3a1=3t
带入an+1=2Sn+1
3t = 2(a1) + 1 => t = 1

即t=1,{an}是等比为3的等比数列

(2)
bn = log3|an+1 = log3|3^n = n

cn = bn/an = n/3^(n-1)

c1 = 1/1
c2 = 2/3^1
c3 = 3/3^2
Tn = 1/1+2/3+3/9+...+n/3^(n-1) (a)
1/3Tn = 1/3+ 2/9+...(n-1)/3^(n-1)+n/3^n (b)
(a)-(b)=>
2/3Tn = 1/1+1/3+1/9+1/3^(n-1) - n/3^n = 1(1-(1/3)^n)/(1-1/3) - n/3^n=3/2 - 3/2*(1/3)^n-n/3^n
=>Tn = 9/4 - 9/4*(1/3)^n-3/2*n/3^n< 9/4

Answer 4