在等比数列中,Sn=5*3^n-1 k求k
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由Sn=3^(n+1)+r可知公比q=3取n=1得a1=9+r取n=2得a1+a2=4a1=27+r解得a1=6,r=-3
a1=s1=m*3+1=3m+1a1+a2=s2=m*3^2+13m+1+a2=9m+1a2=6ma3=s3-s2=m*3^3-m*3^2=18m公比=a3/a2=18m/6m=3a2/a1=36m/
设公比为q,当q=-1时,等比数列{an}的各项是a,-a,a,-a,a,-a…的形式,a≠0.又已知Sn是实数等比数列{an}前n项和,故当n为偶数时,Sn=0,当n为奇数时,Sn=a,故选D.
因为an,Sn,Sn-1/2成等比数列Sn(平方)=an*(Sn-1/2)由an=Sn-S(n-1)Sn(平方)=(Sn-S(n-1))*(Sn-1/2)化简得S(n-1)*Sn=S(n-1)/2-S
设{an}的公比为q,则a2=2q,a3=2q^2则(a2+1)^2=(a1+1)(a3+1)即(2q+1)^2=3(2q^2+1)解得q=1所以{an}为常数数列Sn=na1=2n
当公比为1时,Sn=n,数列{Sn+12}为数列{n+12}为公差为1的等差数列,不满足题意;当公比不为1时,Sn=1−qn1−q,∴Sn+12=1−qn1−q+12,Sn+1+12=1−qn+11−
因数列{an}为等比,则an=2qn-1,因数列{an+1}也是等比数列,则(an+1+1)2=(an+1)(an+2+1)∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a
等比数列(1)a1=3,q=2,n=6Sn=a1(1-q^n)/(1-q)S6=3*(1-2^6)/(1-2)=3*(2^6-1)=3*63=189(2)an=a1*q^(n-1)1/2=8*(1/2
Sn=a1(1-q^n)/(1-q)S1=a1S2=a1(1+q)S3=a1(1+q+q^2)S2+2=a1(1+q)+2S3+2=a1(1+q+q^2)+2[a1(1+q+q^2)+2]*[a1+2
Sn=3^n+aS(n-1)=3^(n-1)+a两式相减得到an=3^n-3^(n-1)=2*n^(n-1)a1=2根据等比数列求和公式,有Sn=a1(3^n-1)/3-1=3^n-1比较两式,有a=
k=-3由等比前n项和公式得Sn=a1(1-q^n)/(1-q)=-[a1/(1-q)]×q^n+a1/(1-q)前面系数与后面的常数相反,∴k=-3
∵等比数列{an}中,Sn=3n+b,∴a1=31+b=3+b,a2=S2-S1=6,a3=S3-S2=18,∴(3+b)•18=36,∴b=-1.故答案为:-1.
∵Sn=3n+a,∴a1=S1=3+a,∵an=Sn-Sn-1=(3n+a)-(3n-1+a)=2×3n-1,∴a1=2.又∵a1=S1=3+a,∴3+a=2,∴a=-1.∴an=2×3n-1.故答案
因数列{an}为等比,则an=3qn-1,因数列{an+1}也是等比数列,则(an+1+1)2=(an+1)(an+2+1)∴an+12+2an+1=anan+2+an+an+2∴an+an+2=2a
an=a1*q^(n-1)96=a1*2^(n-1)192=a1*2^nSn=(a1-a1q^n)/(1-q)189=(a1-192)/(1-2)189=-a1+192a1=3192=3*2^n64=
a1×(½)ˆ7=1∴a1=2ˆ7=128s8=a1[1-(½)ˆ8]/1-½=255
a1=S1=3+a,a2=s2-s1=9+a-(3+a)=6,a3=S3-S2=27+a-(9+a)=18,因为a1·a3=a2²,即18(3+a)=36,解得a=-1再问:求通项公式再答:
(a2+1)²=(a1+1)(a3+1)a1=2,设an公比q(2q+1)²=3(2q²+1)4q²+4q+1=6q²+32q²-4q+2=
设公比为q,a2²=a1*a3(a2+1)²=(a1+1)(a3+1)因为a1=2所以a2²=2a3(a2+1)²=3(a3+1)解得a2=2a3=2所以sn=