www.1862.net > 已知公比不为1的等比数列{An}的首项A1=2,前n项和...

已知公比不为1的等比数列{An}的首项A1=2,前n项和...

(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...

∵公比不为1的等比数列{an}的前n项和为Sn,a1=1,且4a1,3a2,2a3成等差数列,∴6a2=4a1+2a3 ,即6q=4+2q2,解得 q=2.∴an=2n?1,Sn= 1×(1?2n)1?2=2n?1,Sn2n?1?3= 2n?12n?1?3=2+52n?1?3,故当n=3时,有最大值7.故答案为 7.

Sn=a1(1-q^n)/(1-q^n) =(l-(-2)^8)/(1+2) =-255/3

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(1) a1=1/2 an = (1/2).q^(n-1) S3,S2,S4成等差数列 S3+S4 =2S2 (1/2)q^2( 1+ q) = q q^2( 1+ q) = 2q q( q^2+q -2) =0 q(q+2)(q-1) =0 q=-2 an = (1/2) (-2)^(n-1) (2) let S = 1. 2^(-2) + 2.2^(-1) +.....+n.2^(n-2) (1) 2S = 1. 2^(-1) + ...

(Ⅰ)∵公比大于零的等比数列{an}的前n项和为Sn,且a1=1,S4=5S2,即S4=5S2,q>0,∴1?q41?q=5×1?q21?q,解得q=2,an=2n?1.∵数列{bn}的前n项和为Tn,满足b1=1,Tn=n2bn,n∈N*.∴Tn=n2bnTn?1=(n?1)2bn?1,∴bnbn?1=n?1n+1(n>1),∴bnbn?1?...

1) an=c^(n-1),则a(n-1)=c^(n-2),a(n-2)=c^(n-3) c^(n-1)=[c^(n-2)+c^(n-3)]/2, 因为c^(n-3)不等于0,所以化简为c^2=(c+1)/2,解得c=-1/2或1 (2) 设bn=n*an 当c=1时: bn=n,则Sn=1+2+3+...+n=n(n+1)/2 当c=-1/2时: bn=n*(-1/2)^(n-1) Sn =1+2*(-1/2...

由a1,2a7,3a4成等差数列,得4a7=a1+3a4,即4aq6=a+3aq3.变形得(4q3+1)(q3-1)=0,所以q3=?14,或q3=1(舍去).(1)Tn=a1+a4+a7++a3n-2=1+q3+q6++q3n?3=1?q3n1?q3=45[1?(?14)n];(2)由S612S3=a1(1?q6)1?q12a1(1?q3)1?q=1+q312=11...

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