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)∵a3=8,a6=64,∴a6a3=q3=8,∴q=2∴an=a3?qn?3=2n-----------------------------------------(7分)(2)bn=log2an=log22n=n,∴数列{bn}是以1为首项,1为公差的等差数列∴Sn=n(n+1)2------------(14分)

(1)∵等比数列{an}中,a3=3,a6=24∴q3=a6a3=648=8,∴q=2;∴an=a3qn?3=2n.(2)∵a3=8,a6=64∴b3=8,b5=64,设等差数列{bn}的公差为d,∴d=b5?b35?3=28,∴bn=b3+(n-3)d=28n-76,Sn=b1n+n(n?1)2d=?48n+n(n?1)2×28=14n2-62n

设等比数列{an}的公比为q,因为a3+a6=36,①a4+a7=18 ②,②①可得a4+a7a3+a6=q=12,故a3+a6=a1q2+a1q5=14a1+132a1=36,解得a1=27,故通项公式an=27×(12)n?1=28-n,令28-n=12=2-1,解得n=9

解(1)设数列前6项的公差为d,d为整数,则a5=-1+2d,a6=-1+3d,d为整数,又a5,a6,a7成等比数列,所以(3d-1)2=4(2d-1),解得d=1,-------4分当n≤4时,an=n-4,由此a5=1,a6=2,数列第5项起构成以2为公比的等比数列.当n≥5时,an=2n-5,故...

(1)∵{an}为等比数列,a3=4,a6=32,设公比为q,∴a1q2=4a1q5=32,解得a1=1,q=2,∴an=2n?1,Sn=1?2n1?2=2n-1.(2)∵Sn=2n?1,∴T=Sn+64Sn+1=2n-1+642n≥22n?642n-1=15.当且仅当2n=642n,即n=3时,T取最小值15.

(1)设公差为d,则∵S4=14,且a1,a3,a7成等比数列∴4a1+6d=14,(a1+2d)2=a1(a1+6d)∵d≠0,∴d=1,a1=2,∴an=n+1,sn=n(2+n+1)2=n(n+3)2.(2)1an?an+1=1(n+1)(n+2)=1n+1-1n+2 ∴Tn=12?13+13?14+…+1n+1?1n+2=12-1n+2=n2(n+2)∵Tn≤λan+1对任意的...

(Ⅰ)设公差为d,∵Sn=14,且a1,a3,a7成等比数列,∴4a1+6d=14(a1+2d)2=a1(a1+6d),…(4分)解得d=0(舍)或d=1,所以a1=2,故an=n+1.…(7分)(Ⅱ)∵an=n+1,∴1anan+1=1(n+1)(n+2)=1n+1?1n+2,所以Tn=12?13+13?14+…+1n+1-1n+2=12?1n+2,…(1...

(Ⅰ)由条件知a2-a3=2(a3-a4).(2分)即a1q-a1q2=2(a1q2-a1q3),又a1?q≠0.∴1-q=2(q-q2)=2q(1-q),又q≠1.∴q=12.(4分)∴an=64?(12)n?1=(12)n-7.(6分)(Ⅱ)bn=log2an=7-n.{bn}前n项和Sn=n(13?n)2.∴当1≤n≤7时,bn≥0,∴Tn=Sn...

a5/a3=q²所以q=±2,所以a1=1再了利用等比数列求和公式就可以了,所以应该有两个解吧

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