一直数列an满足:a1=1/2,3(1+a n+1)/1-an=2(1+an)/1-a n+1,ana n+1/1),数
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一直数列an满足:a1=1/2,3(1+a n+1)/1-an=2(1+an)/1-a n+1,ana n+1/1),数列bn满足bn=(a n+1)^2(n>/1) (1)求an bn通项公式 (2)证:数列bn中任意三项不可能成等差数列
![一直数列an满足:a1=1/2,3(1+a n+1)/1-an=2(1+an)/1-a n+1,ana n+1/1),数](/uploads/image/z/15244682-50-2.jpg?t=%E4%B8%80%E7%9B%B4%E6%95%B0%E5%88%97an%E6%BB%A1%E8%B6%B3%3Aa1%3D1%2F2%2C3%281%2Ba+n%2B1%29%2F1-an%3D2%281%2Ban%29%2F1-a+n%2B1%2Cana+n%2B1%2F1%29%2C%E6%95%B0)
(1).证明:a2-2abcos(60°+C)=a2-2ab(cos60°cosC-sin60°sinC)=a2-abcosC+√3absinC——(1)
根据余弦定理得:c2=a2+b2-2abcosC→-abcosC=(c2-a2-b2)/2——(2)
根据正弦定理得:a/sinA=c/sinC→asinC=csinA——(3)
将(2)和(3)代入(1)中得:a2-abcosC+√3absinC=(c2+a2-b2)/2+√3bcsinA
=c2-bccosA+√3bcsinA
=c2-2bccos(60°+A)
故:a2-2abcos(60°+C)=c2-2bccos(60°+A)
(2)证明:(a-b)2cos2(C/2)+(a+b)2sin2(C/2)=(a-b)2(1+cosC)/2+(a+b)2(1-cosC)/2
= a2+b2-2abcosC= c2
故:(a-b)2cos2(C/2)+(a+b)2sin2(C/2)=c2
根据余弦定理得:c2=a2+b2-2abcosC→-abcosC=(c2-a2-b2)/2——(2)
根据正弦定理得:a/sinA=c/sinC→asinC=csinA——(3)
将(2)和(3)代入(1)中得:a2-abcosC+√3absinC=(c2+a2-b2)/2+√3bcsinA
=c2-bccosA+√3bcsinA
=c2-2bccos(60°+A)
故:a2-2abcos(60°+C)=c2-2bccos(60°+A)
(2)证明:(a-b)2cos2(C/2)+(a+b)2sin2(C/2)=(a-b)2(1+cosC)/2+(a+b)2(1-cosC)/2
= a2+b2-2abcosC= c2
故:(a-b)2cos2(C/2)+(a+b)2sin2(C/2)=c2
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