lstd.net
当前位置:首页 >> 已知等比数列{An}满足2A1+A3=3A2,且A3+2是A2与A4的等差中项;(1)求数列{An}... >>

已知等比数列{An}满足2A1+A3=3A2,且A3+2是A2与A4的等差中项;(1)求数列{An}...

解:设{an}公比为q(1)2a1+a3=3a2,2a1+a1q=3a1qq-3q+2=0,(q-1)(q-2)=0q=1或q=2a3+2是a2、a4的等差中项,则2(a3+2)=a2+a42a3+4=a2+a4,若q=1,则a2+a4=2a3,等式变为4=0,等式不成立,因此q≠1q=22a1q+4=a1q+a1q,q=2代

(Ⅰ)设等比数列{an}的首项为a1,公比为q,依题意,∵2a1+a3=3a2,且a3+2是a2,a4的等差中项 ∴ a1(2+q2)=3a1q① a1(q+q3)=2a1q2+4② 由 ①得 q2-3q+2=0,解得q=1或q=2. 当q=1时,不合题意舍;当q=2时,代入(2)得a1=2,所以an=2n.

解:设{an}公比为q(1)2a1+a3=3a2,2a1+a1q=3a1qq-3q+2=0,(q-1)(q-2)=0q=1或q=2a3+2是a2、a4的等差中项,则2(a3+2)=a2+a42a3+4=a2+a4,若q=1,则a2+a4=2a3,等式变为4=0,等式不成立,因此q≠1q=22a1q+4=a1q+a1q,q=2代

因为an=2^n,所以 log2 1/an(2为角标)= -n所以 bn=2^n-nSn=2-1+2^2-2+2^3-3++2^n-n =(2+2^2+2^3++2^n)-(1+2+3++n)= 2^(n+1)-2-(1+n)*n/2Sn-2^(n+1)+47=2^(n+1)-2-(1+n)*n/2-2^(n+1)+47=45-(1+n)*n/290(n+1

(Ⅰ)设等比数列{an}的公比为q,依题意,有2a1+a3=3a2a2+a4=2(a3+2),即a1(2+q2)=3a1q,①a1(q+q3)=2a1q2+4,②,由①得q2-3q+2=0,解得q=2或q=1,当q=1时,不合题意;当q=2时,代入②得a1=2,∴an=22n-1=2n;(Ⅱ)bn=an+log21an=2n+log212n=2n-n,∴Sn=2-1+22-2+23-3+…+2n-n=(2+22+23+…+2n)-(1+2+3+…+n)=2(12n)12-n(n+1)2=2n+1-2-12n-12n2.

an=2^nbn=an-nSn=S(an)-n(n+1)/2=2(2^n-1)-n(n+1)/2=2^(n+1)-2-n(n+1)/2Sn-2^(n+1)+47=45-n(n+1)/2<0得n(n+1)>90得n最小为10

第一问我已经求出:an=2^n .你求错了照你说的a1=2a2=4a3=8你觉得满足a1+a3=3a2吗?

(Ⅰ)设等比数列{an}的首项为a1,公比为q,以题意,有2a1+a3=3a2a2+a4=2(a3+2),即a1(2+q2)=3a1q (1)a1(q+q3)=2a1q2+4 (2)由(1)得:q2-3q+2=0,解得q=1或q=2.当q

(1)∵等比数列{an}满足a1a2=2a3,且a1,a2+2,a3成等差数列,∴a12q=2a1q22(a1q+2)=a1+a1q2,解得a1=4q=2,∴a n=2n+1.∵b1log2a1+b2log2a2+…+bnlog2an=n(n+1)

由题设:令n=1得:λa1a1=S1+S1=2a1 则a1=0或a1=2/λ若a1=0,则Sn=0,从而an=0若a1=2/λ,由已知:λa1a(n+1)=S1+S(n+1)两式相减得:λa1[a(n+1)-an]=S(n+1)-Sn=a(n+1) 即2[a(n+1)-an]=a(n+1)所以a(n+1)=2an 说明{an}是

相关文档
sgdd.net | ncry.net | ymjm.net | xcxd.net | gtbt.net | 网站首页 | 网站地图
All rights reserved Powered by www.lstd.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com