www.1862.net > 函数是等比数列,4s1,3s2,2s3成等差数列,且s4=15,...

函数是等比数列,4s1,3s2,2s3成等差数列,且s4=15,...

设函数an=a1*q^(n-1),又4s1,3s2,2s3成等差数列,那么有 2*3s2=4s1+2s3,6*a1(1+q)=4a1+2*a1*(1+q+q^2),解出q=2或q=0(舍去) s4=15,s4=a1*q^3=15,所有a1=15/8,通项公式是:an=15/8*2^(n-1)=15*2^(n-4)

设等比数列{an}的公比为q,∵S1,2S2,3S3成等差数列∴4S2=S1+3S3,即4(a1+a2)=a1+3(a1+a2+a3)∴a2=3a3,即q=13又S4=4027∴a1(1?q4)1?q=4027解得a1=1∴an=(13)n?1

(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)...

S2,S4,S3成为等差数列,∴S2+S3=2xS4

(1)解:设数列{an}的公比为q,∵且S3,S2,S4成等差数列,∴S3+S4=2S2,即(a1+a2+a3)+(a1+a2+a3+a4)=2(a1+a2)∴2a3+a4=0,q=a4a3=-2,∴an=a1qn-1=(-2)n-1;(2)证明:|an|=2n-1,bn=log22n?1=n?1,∴bn+1|an|=n2n?1,∴Tn=120+221+322...

(1)∵公比不为1的等比数列{an}的首项a1=12,前n项和为Sn,且a3+S5,a4+S4,a5+S3成等差数列,∴a3+S4-a3-S3=a5+S5-a4-S4,∴2a5-3a4+a3=0,∴2q2-3q+1=0,∵q≠1,∴q=12,∴an=12n.(2)bn=an+an+12?3n=34?(32)n,∵cn=3n4bn,∴cn=n?(23)n,∴Tn=1...

:(Ⅰ)设正项等比数列{an}(n∈N*)的公比为q(q>0),又a1=12,∴an=12?qn-1,∵S3+a3、S5+a5、S4+a4成等差数列,∴2(S5+a5)=(S3+a3)+(S4+a4),即2(a1+a2+a3+a4+2a5)=(a1+a2+2a3)+(a1+a2+a3+2a4),化简得4a5=a3,∴4a1q4=a1q2,化为4q...

s3=?

解:等差数列an,a1=1,设公差为d, 由 得:S3=3+3d,S4=4+6d,S5=5+10d ∵S3+S4=S5 ∴3+3d+4+6d=5+10d,解得d=2 ∴an通项公式为:an=1+2(n-1)=2n-1

(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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