已知数列{an}的前n项和Sn=n(n+1)/2 (n∈N*) (1)求数列{an}的通项公式 (2)设数列{bn}满足(2an-1)(2^(bn)-1)

Question

已知数列{an}的前n项和Sn=n(n+1)/2 (n∈N*) (1)求数列{an}的通项公式 (2)设数列{bn}满足(2an-1)(2^(bn)-1)=1,Tn为{bn}的前n项和，求证：
2Tn>log2(2an+1),n∈N*.

Answer 1

(1)
a(n) = S(n) - S(n-1)
a(n) = n(n+1)/2 - (n-1)n/2 = n

(2)
(2an-1)(2^(bn)-1) = 1
(2n - 1)(2^(bn)-1) = 1
2^(bn)-1 = 1/(2n - 1)
2^(bn) = 2n/(2n - 1)
b(n) = log(2)[2n/(2n - 1)]
T(n) = b(1) + b(2) +...+ b(n)
T(n) = log(2){2/1}*{4/3}*...*{2n/(2n - 1)}
2T(n) = log(2){2/1}*{4/3}*...*{2n/(2n - 1)}^2
log2(2an+1) = log2(2n+1),//是这样的意思么
//这里的log是不是指以2为底数的对数
如果是
因为(a-1)(b-1) = ab -(a+b)+1>0 ab>a+b-1>1
{2/1}*{4/3}*...*{2n/(2n - 1)}^2>[4-1]...[4n/(2n-1) - 1] = 3*...*[(2n+1)/(2n-1)]
= [3/1 * 5/3 * ... * (2n-1)/(2n-3) * (2n+1)/(2n-1)]
= [2n+1] = (2an+1)
log(2)x是增函数，所以
2Tn>log2(2an+1),n∈N*.

Answer 2

http://u.youku.com/user_show/uid_%e6%9f%93%e8%89%b2%e7%9a%84%e9%9d%92%e6%98%a5