www.1862.net > 等比数列{An}的前n项和为sn,已知s4=1,s8=17,求{An}...

等比数列{An}的前n项和为sn,已知s4=1,s8=17,求{An}...

设等比数列{an}的公比为q,由S4=1,S8=17,得S8-S4=17-1=16,∴q4=S8?S4S4=16,解得q=±2,当q=2时,由S4=a1(1?q4)1?q=15a1=1可得a1=115,∴Sn=a1(1?qn)1?q=115(2n-1);当q=-2时,由S4=a1(1?q4)1?q=-5a1=1可得a1=?15,∴Sn=a1(1?qn)1?q=115[(-2)...

不用公式进行计算, 因为等差数列有个规律: Sn,S2n-Sn,S3n-S2n。。。 是等差数列, 因为S4\S8=1\3, 那么S4=k,S8-S4=2k, S12-S8=3k,S16-S12=4k, 所以S8\S16=3k\10k=3\10。

s8=a1(1-q^8)/(1-q) =a1(1-q^4)(1+q^4)/(1-q) =(1+q^4)*[a1(1-q^4)/(1-q)] =(1+q^4)*S4 =(1+2^4)*1 =1+16 =17 希望帮到你 望采纳 谢谢 加油

在等比数列{an}中,设公比q(q≠0),前n项和为Sn,当a1=1时,有4a1,2a2,a3成等差数列,∴4a2=4a1+a3,即4q=4+q2,∴q=2∴S4=1×(1?24)1?2=15.故答案为:15.

所求式子为S13-S12+S14-S13+S15-S14+S16-S15=S16-S12,根据S4和S8,有a1(1-q4)/(1-q)=1,a1(1-q8)/(1-q)=4,等式两边相除有q4=3.a1/(1-q)=-1/2. 则S16-S12=a1(1-q16)/(1-q)-a1(1-q12)/(1-q)=a1(q12-q16)/(1-q)=-1/2*(3^3-3^4)=27。(先判断比值不...

s3=?

(Ⅰ)由题意得a1+2d=3,4a1+6d=10,解得a1=1,d=1,从而数列{an}的通项公式为an=n.(Ⅱ)bn=1anan+1=1n(n+1)=1n?1n+1,∴Tn=(1-12)+(12?13)+…+(1n?1n+1)=nn+1.

a1=S1=1 4a1+4x3d/2=4(2a1+2xd/2) d=2 6a1+6x5d/2=m(4a1+4x3d/2) m=9/4

(1)∵a4+S4,a5+S5,a6+S6成等差数列,∴a5+S5-(a4+S4)=a6+S6-(a5+S5),∴2a5-a4=2a6-a5,∴2a6-3a5+a4=0.∵数列{an}为等比数列,∴a5=a4q,a6=2a4q2,∴2q2-3q+1=0,∴(2q-1)(q-1)=0.∵数列{an}公比不为1,∴q=12.∴an=2×(12)n?1,∴an=(12)...

(1)∵a4+S4,a5+S5,a6+S6成等差数列,∴2(a5+S5)=(a4+S4)+(a6+S6)…(2分)即2a6-3a5+a4=0,∴2q2-3q+1=0,∵q≠1,∴q=12,…(4分)所以等比数列{an}的通项公式为an=12n;…(6分)(2)bn=an+an+12?3n=34?(32)n,…(9分)∴数列{bn}为等比...

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