一等比数列『an』的前三次项依次为a,2a 2
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S(n+1)=4an+2,所以:Sn=4a(n-1)+2两式相减:a(n+1)=4an-4a(n-1)a(n+1)-2an=2[an-2a(n-1)]令:cn=an-2a(n-1),c2=a2-2a1
Sn=a1(1-q^n)/(1-q)a1=1q=-2n=8Sn=1*(1-(-2)^8)/(1-(-2))=-85
设数列An的公比为q则:An=(a1)q^(n-1)而:a10^2=a15所以:((a1)q^(10-1))^2=(a1)q^(15-1)q^4=1/a1因q>1,因此:a1>0设另有数列Bn,Bn=
(1)(a-1)+(a平方-2)+...+(a的n次方-n)=(a+a平方+…..+a的n次方)-(1+2+….+n)=a[a的n次方-1]/(a-1)-n(n+1)/2(2)(2-3×5的-1次方)
a(n)=a+(n-1)d,s(n)=na+n(n-1)d/2.[a(4)]^2=a(1)a(13)=[a+3d]^2=a[a+12d],a^2+6ad+9d^2=a^2+12ad,0=9d^2-6a
列一下{an+1}等比的递推式易得q=1Sn=2n
由a6=a1*q5得q=-1/2∴(a1-anq)/(1-q)=255/64截得解得an=-3/512∴q=10
n=1时,a1=1+3a1.即a1=-1/2.n>1时,an=Sn-Sn-1=1+3an-(1+3a(n-1))=3an-3a(n-1),即an=3/2a(n-1),即an=-1/2*(3/2)^(n
S9-S3=a9+a8+a7+a6+a5+a4S6-S9=-a9-a8-a7因S3,S9,S6,成等差数列则a9+a8+a7+a6+a5+a4=-a9-a8-a7a6+a5+a4=-2a9-2a8-2
Sn=2an-3n+5S(n-1)=2a(n-1)-3(n-1)+5相减an=2a(n-1)+3an+3=2a(n-1)+6an+3=2[2a(n-1)+3]
Sn-S(n-1)=an=2an+3-2a(n-1)-3=2an-2a(n-1)an=2a(n-1){an}为等比数列,公比为2
a1,a7,a4成等差数列2a7=a1+a42a1q^6=a1+a1q^32q^6=1+q^32q^6-q^3-1=(2q^3+1)(q^3-1)=0因为公比Q不等于1,所以,q^3=-1/2,2S3
Sn=4An-3S(n-1)=4A(n-1)-3Sn-S(n-1)=An=4An-3-[4A(n-1)-3]=4an-3-4A(n-1)+3=4An-4A(n-1)3An=4A(n-1)An/A(n-
an=Sn-S(n-1)=2(an-3)-2[a(n-1)]-3=2an-2a(n-1)]an=2a(n-1)所以an是等比数列q=1S1=a1所以a1=2(a1-3)a1=6所以an=6*2^(n-
设等比数列{an}的公比为q,则可得an=2•qn-1,故an+1=2•qn-1+1,可得a1+1=3,a2+1=2q+1,a3+1=2q2+1,由于数列{an+1}也是等比数列,故(2q+1)2=3
1.a1=2,a2=4,a3=8,a4=16S8=5102.a1=16,a2=8,a4=4,a4=2S8=255/8再问:咋算出来的啊再答:a1+a18q^3=18a1q+a2q^2=12相除(1+q
解题思路:利用等比数列性质求出公比,分类讨论,注意n是正整数解题过程:最终答案:5
(1)令n=1,得a1=-1.Sn=2an+n,S(n+1)=2a(n+1)+n+1.两式相减,得a(n+1)=2a(n+1)-2an+1.整理得a(n+1)-1=2(an-1),a1-1=-2.综上
(1)令S=a1+a2+.+an,即S=a1+a1*q+.+a1*q^(n-1)则qS=a1*q+a1*q^2+a1*q^n故(1-q)S=a1-a1*q^n得S=a1(1-q^n)/(1-q)(2)