Question

已知函数f(x)=(2x+3)/3x,数列{an}满足a1=1,an+1=f(1/an)...

已知函数f(x)=(2x+3)/3x,数列{an}满足a1=1,an+1=f(1/an)，n∈N*.

(1)[我已经做好 an=(2n+1)/3]

（2）令Tn=a1a2-a2a3+a3a4-a4a5a+...-a2na2n+1,求a2n-1 - a2n+1及Tn

（3）令bn=1/(an-1*an),(n>=2),b1=1,Sn=b1+b2+...bn,若Sn<(m-2004)/2对一切n∈N*成立,求最小正整数m.

好人有好报..................

Answer 1

分闲少啊

Answer 2

（2）：a(2n-1)-a(2n+1)=(4n-1)/3-(4n+5)/3=—4/3
a2n=(4n+2)/3 所以Tn=—4/3[(4*1+2)/3+(4*2+2)/3+(4*3+2)/3+……+(4n+2)/3]
=—4/3*[（2n+4)n]/3
=—n(8n+16)/9