www.1862.net > 已知等比数列{An}中,A3=4,A7=64.(1)求数列{An...

已知等比数列{An}中,A3=4,A7=64.(1)求数列{An...

(1)设等比数列{an}的公比为q,∵a3=4,a7=64,∴a1q2=4a1q6=64,解得a1=1,q=2.∴an=1×2n?1=2n?1.(2)Sn=1×(2n?1)2?1=2n-1.

∵等比数列{an}中,a3=4,a7=64,∴a1q2=4,a1q6=64,∴q=±2,a1=1,∴S7=a1(1?q7)1?q=1?271?2或1?(?2)71+2=127或43.故答案为:127或43.

(1)a4=a1*q^3 64=(-1)*q^3 q^3=-64=(-4)^3 q=-4 S4=a1*(q^4-1)/(q-1) =(-1)*[(-4)^4-1]/(-4-1) =51 (2)a3=a1q^2=3/2 (1) s3=a1+a1q+a1q^2=9/2 (2) 由 (1)得:a1=3/(2q^2) [3/(2q^2)](1+q)=9/2-3/2=3 -2q^2+q+1=0 q=1,或q=-1/2 代入得a1=3/2...

a1×a9=64,所以a3×a7=64,又a3+a7=20,解得a3=4,a7=16;或a3=16,a7=4.当a3=4,a7=16时,a11=64.当a3=16,a7=4时,a11=1

(1)设数列{an}的公比为q.由等比数列性质可知:a1a7=a3a5=64,而a1+a7=65,an+1<an.∴a1=64,a7=1,(3 分)由64q6=1,得q=12,或q=-12(舍),(5 分)故an=27?n.(7 分)(2)等比数列{an}中,∵a1=64,q=12,∴S5=64×[1?(12)5]1?12=124...

(1)在等比数列中,由a1+a7=65,a3?a5=64,得a1+a7=65,a3?a5=a1a7=64,解得a1=1,a7=64或a1=64,a7=1,由an+1<an得数列为递减数列,∴a1=64,a7=1,解得64q6=1,即q6=164=(12)6,解得q=12或q=?12(舍去).∴求数列{an}的通项公式问an=64?(...

(1)q^3=a7/a4=8/2=4 q=4^(1/3)=2^(2/3) a1=a4/q^3=2/4=1/2 an=a1*q^(n-1)=1/2*(2^(2/3))^(n-1)=2^(-1+2n/3-2/3)=2^(2n/3-5/3) (2) a2+a5=a2(1+q^3)=18 a3+a6=a3(1+q^3)=9 下式/上式得:q=a3/a2=1/2 a2+a5=a1*1/2+a1*(1/2)^4=18 a1=32 an=a1*q...

(Ⅰ)设等比数列{an}的公比为q(q≠0),由a7=a1q6=1,得a1=q-6,从而a4=a1q3=q-3,a5=a1q4=q-2,a6=a1q5=q-1.…(3分)因为a4,a5+1,a6成等差数列,所以a4+a6=2(a5+1),即q-3+q-1=2(q-2+1),q-1(q-2+1)=2(q-2+1).所以q=12.故an=a1qn-...

设数列前6项的公差为d,d为整数,由a3=-1,得:a5=a3+2d=-1+2d,a6=a3+3d=-1+3d,又a5,a6,a7成等比数列,且a7=4,所以(3d-1)2=4(2d-1),解得d=59或d=1,因为d为整数,所以d=1.所以,当n≤6时,an=a3+(n-3)×1=-1+(n-3)=n-4,由此a5=1...

(I)设等比数列{an}的公比为q.由a1a3=4可得a22=4,(1分)因为an>0,所以a2=2(2分)依题意有a2+a4=2(a3+1),得2a3=a4=a3q(3分)因为a3>0,所以,q=2..(4分)所以数列{an}通项为an=2n-1(6分)(II)bn=an+1+log2an=2n+n-1(18分)可得S...

网站地图

All rights reserved Powered by www.1862.net

copyright ©right 2010-2021。
www.1862.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com