已知函数f(x)=2x+3/3x,数列{an}满足a1=1,an+1=f(1/an),n属于N (1)求数列{an}通项公式,

题目

已知函数f(x)=2x+3/3x,数列{an}满足a1=1,an+1=f(1/an),n属于N (1)求数列{an}通项公式,
(2)令Tn=a1a2-a2a3+a3a4-a4a5+…+a2n-1a2n-a2na2n+1,求Tn

答案

a(n+1)=(2/an+3)/3/an=(2+3an)/3=2/3+ana(n+1)-an=2/3等差数列an=a1+2/3 *(n-1)=2n/3+1/3a1a2=1*5/3=3/3*5/3=3*5/9a2a3=5/3*7/3=5*7/9a1a2-a2a3=4*5/9=4*(4+1)/9 (1)a3a4=7/3*9/3=7*9/9a4a5=9/3*11/3=9*11/9a3a4-a4a...