已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn}的
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![已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn}的](/uploads/image/z/2778185-65-5.jpg?t=%E5%B7%B2%E7%9F%A5%EF%BD%9Ban%EF%BD%9D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E5%85%AC%E6%AF%94q%3D%E6%A0%B9%E5%8F%B72%2CSn%E4%B8%BA%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C.Tn%3D%EF%BC%8817Sn-S2n%EF%BC%89%2Fan%2B1%2Cn%E5%B1%9E%E4%BA%8EN%2A%2C%E8%AE%BETn0+%E4%B8%BA%EF%BD%9BTn%E5%B7%B2%E7%9F%A5%EF%BD%9Ban%EF%BD%9D%E4%B8%BA%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E5%85%AC%E6%AF%94q%3D%E6%A0%B9%E5%8F%B72%2CSn%E4%B8%BA%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C.Tn%3D%EF%BC%8817Sn-S2n%EF%BC%89%2Fan%2B1%2Cn%E5%B1%9E%E4%BA%8EN%2A%2C%E8%AE%BETn0+%E4%B8%BA%EF%BD%9BTn%EF%BD%9D%E7%9A%84)
已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn}的
已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn
已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn}的最大项,则n0为多少
已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn已知{an}为等比数列,公比q=根号2,Sn为{an}的前n项和.Tn=(17Sn-S2n)/an+1,n属于N*,设Tn0 为{Tn}的
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设{an}为等比数列,公比q=根号2,Sn为{an}前n项和
可得 a(n+1)=a1*2^(n/2)
Sn=a1*[1-2^(n/2)]/(1-√2)
S2n=a1*[1-2^n]/(1-√2)
Tn=(17Sn-S2n)/a(n+1)
化简后
=[16-17*2^(n/2)+2^n]/(1-√2)*2^(n/2)
=-(√2+1)(16/...
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设{an}为等比数列,公比q=根号2,Sn为{an}前n项和
可得 a(n+1)=a1*2^(n/2)
Sn=a1*[1-2^(n/2)]/(1-√2)
S2n=a1*[1-2^n]/(1-√2)
Tn=(17Sn-S2n)/a(n+1)
化简后
=[16-17*2^(n/2)+2^n]/(1-√2)*2^(n/2)
=-(√2+1)(16/2^(n/2)-17+2^(n/2))
由均值不等式
16/2^(n/2)-17+2^(n/2)≥-9 (n=4时等号成立)
故原式=-(√2+1)(16/2^(n/2)-17+2^(n/2))
≤9(√2+1)
Tn0=9(√2+1) n0=4
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