已知函数f(x)=2x+3/3x,数列{an}满足a1=1,an+1=f(1/an),n为正整数
问题描述:
已知函数f(x)=2x+3/3x,数列{an}满足a1=1,an+1=f(1/an),n为正整数
(2)令Tn=a1a2-a2a3+a3a4-a4 a5+……-a2n a2n+1,求a2n-1-a2n+1及Tn
答
A(n+1)=f(1/An)A(n+1)=An(2/An +3)/3=2/3 +AnA(n+1)-An=2/3Tn=A2(A1-A3)+A4(A3-A5)+...+A(2n)(A(2n-1)-A(2n+1))=A2(-2d)+2A2(-2d)+..nA2(-2d)=A2(-2d)(1+2+3+4+...+n)