求数列{an}{bn}满足a1=1,a2=r,r>0,bn=ana(n+1)且{bn}是公比为q的等比,设Cn=a (2

求数列{an}{bn}满足a1=1,a2=r,r>0,bn=ana(n+1)且{bn}是公比为q的等比,设Cn=a (2n-1)+a2n
(1)求{an}通项
(2)设dn=lgC(n+1)/lgCn,求{dn}最大项与最小项的值
r=2的19.2次方-1,q=1/2
伤心不已的我 1年前 已收到1个回答 举报

sdxtxy 幼苗

共回答了12个问题采纳率:83.3% 举报

1 = a1a2 = r,故bn = r*q^(n-1)
又b(n+1)/bn = a(n+1)*a(n+2)/(an*a(n+1)) = a(n+2)/an、b(n+1)/bn = q
可得当n为奇数时an = a1*q^((n+1)/2 - 1) = q^((n-1)/2)
当n为偶数时an = a2*q^(n/2 - 1) = r*q^(n/2 - 1)
cn = a(2n-1)+a2n = q^(n-1) + r*q^(n-1) = (1+r)*q^(n-1)

1年前

10
可能相似的问题
Copyright © 2024 YULUCN.COM - 雨露学习互助 - 16 q. 0.045 s. - webmaster@yulucn.com