f(x)=(x+3)/(x+1)(x≠-1),数列|bn|满足bn=|an-√3|
函数数列综合问题:
f(x)=(x+3)/(x+1)(x≠-1),数列{an}满足:a1=1,an+1=f(an),
数列|bn|满足bn=|an-√3|,sn=b1+……+bn,(n属于非0的全体自然数)
用数学归纳法证明:bn小于或等于《=[(√3-1)^n]/【2^(n-1)】
人气:437 ℃ 时间:2020-03-30 09:52:31
解答
an+1-√3=(an+3/an+1)-√3=……=(1-√3)*(an-√3)/(an+1)
又bn=an-√3绝对值,所以得bn+1=(1-√3)/【(an)+1】的绝对值*bn
∵an=1,可推导出an>=1恒成立,进一步推导得,bn+1
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