已知f(x)=根号下4+1/x2,数列{an}的前n项和为Sn,点Pn(an,1/an+1)(n属于N*)在曲线y=f(x)上,且a1=1,an>0.求a
已知f(x)=根号下4+1/x2,数列{an}的前n项和为Sn,点Pn(an,1/an+1)(n属于N*)在曲线y=f(x)上，且a1=1,an>0.求an 通项
人气：388 ℃ 时间：2020-03-30 07:16:00
解答
1/an+1=√(4+1/an^2)a2=√5/51/(an+1)^2=4+1/an^21/(an+1)^2-1/an^2=4所以Tn=1/(an+1)^2是等差数列T1=5Tn=5+(n-1)*4=1/(an+1)^2=4n+1所以an+1=√[1/(4n+1)]an=√[1/(4n-3)] (n>=1)
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