等差数列{an}中,a1=1,前n项和为Sn,等比数列{bn}各项均为正数,b1=2且s2+b2=7,s4-b3=2.求an与bn设Cn=a2n-1/a2n,Tn=c1c2c3...cn,求证Tn>=1/(2根号n)

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等差数列{an}中,a1=1,前n项和为Sn,等比数列{bn}各项均为正数,b1=2且s2+b2=7,s4-b3=2.求an与bn设Cn=a2n-1/a2n,Tn=c1c2c3...cn,求证Tn>=1/(2根号n)
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等差数列{an}中,a1=1,前n项和为Sn,等比数列{bn}各项均为正数,b1=2且s2+b2=7,s4-b3=2.求an与bn设Cn=a2n-1/a2n,Tn=c1c2c3...cn,求证Tn>=1/(2根号n)
等差数列{an}中,a1=1,前n项和为Sn,等比数列{bn}各项均为正数,b1=2且s2+b2=7,s4-b3=2.求an与bn
设Cn=a2n-1/a2n,Tn=c1c2c3...cn,求证Tn>=1/(2根号n)

等差数列{an}中,a1=1,前n项和为Sn,等比数列{bn}各项均为正数,b1=2且s2+b2=7,s4-b3=2.求an与bn设Cn=a2n-1/a2n,Tn=c1c2c3...cn,求证Tn>=1/(2根号n)
设{an}公差为d,数列{bn}公比为q,数列各项均为正,又b1=2>0,因此q>0
S2+b2=7
2a1+d+b1q=7
2+d+2q=7
d+2q=5 (1)
S4-b3=2
4a1+6d-b1q²=2
4+6d-2q²=2
6d-2q²=-2 (2)
由(1)得d=5-2q,代入(2)
6×(5-2q)-2q²=-2
整理,得
q²+6q-16=0
(q+8)(q-2)=0
q=-8(