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下面一道有趣的数列大题,大家有空看下吧:数列{an}恒满足等式a(n+1)=1/2an+√3/2×√(1-a²n),其中a1∈(1/2,1)数列{bn}恒满足等式bn=an/(2^n)求证:1.a(n+2)=an2.设数列{bn}的前n项和为Tn,则T
题目详情
下面一道有趣的数列大题,大家有空看下吧:
数列{an}恒满足等式 a(n+1) = 1/2 an + √3 /2 × √(1-a²n) ,其中a1∈(1/2 ,1)
数列{bn}恒满足等式bn= an / (2^n)
求证:
1.a(n+2)=an
2.设数列{bn}的前n项和为Tn,则Tn<√7 /3
数列{an}恒满足等式 a(n+1) = 1/2 an + √3 /2 × √(1-a²n) ,其中a1∈(1/2 ,1)
数列{bn}恒满足等式bn= an / (2^n)
求证:
1.a(n+2)=an
2.设数列{bn}的前n项和为Tn,则Tn<√7 /3
▼优质解答
答案和解析
(1)、令a1=sinα α∈(π/3,π/2)
a2=sinα*cos(π/3)+sin(π/3)*|cosα|
=sin(α+π/3)
a3=sin(α+π/3)*cos(π/3)+sin(π/3)*|cos(α+π/3)|
α∈(π/3,π/2),则 α+π/3∈(2π/3,7π/6)
cos(α+π/3)
a2=sinα*cos(π/3)+sin(π/3)*|cosα|
=sin(α+π/3)
a3=sin(α+π/3)*cos(π/3)+sin(π/3)*|cos(α+π/3)|
α∈(π/3,π/2),则 α+π/3∈(2π/3,7π/6)
cos(α+π/3)
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