在等比数列{an}中,设sn为其n项和,且a1>0,s3=s11
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(1)因为an是等比数列又a7^2=a9a1^2*q^12=a1*q^8a1*q^4=1即a5=1因为q^4>0所以a1>0因为a8>a9a1*q^7>a1*q^8所以q^8
a7²=a9=a7q^2,a7≠0,∴q^2=a7,a8>a9,∴a8(1-q)>0,条件不足
(1)由等比数列的性质可知,a9=q^2a7而a7∧2=a9,所以a7(a7-q^2)=0,等比数列的任意一项不能为0,所以a7=q^2a8=q^3,a9=q^4,a8>a9=>q^4-a^3
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
Sn,S2n-Sn,S3n-S2n成等比数列48,12,3S3n-S2n=3S3n=3+S2n=63
a1+an=66,a2an-1=128=a1an两式解得a1=2,an=64或a1=64,a1=2Sn=(a1-an*q)/(1-q)前者的话解得q=2后者的话解得q=1/2再带回去得n=6(两种情况
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
S10=a1(1+q+q^2+.+q^9)S5=a1(1+q+.+q^4)故S10/S5=a1(1+q+q^2+.+q^9)/a1(1+q+.+q^4)=[(1+q+.+q^4)+(1+q+.+q^4
Sn=a1(1-q^n)/(1-q)Sn+1=a1[1-q^(n+1)]/(1-q)Sn+2=a1[1-q^(n+2)]/(1-q)2Sn+2=Sn+Sn+1a1[1-q^(n+1)]/(1-q)+a
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
(1)an=x^(n-1)【等比数列通项公式】 当x≠1时sn=(1-x^n)/(1-x)【等比数列求和公式】 于是bn=[(x^n)-x^(n-1)]/[(x^n)-1] 当x=1时sn=n,
设公比是q,S10=(a1+a2+a3+a4+a5)+(a6+a7+a8+a9+a10)其中(a6+a7+a8+a9+a10)=q^5*(a1+a2+a3+a4+a5)所以S10=(a1+a2+a3+
n=1时,a1=S1=2a1+3-7,∴a1=4n>1时,Sn=2an+3n-7①,S(n-1)=2a(n-1)+3(n-1)-7②①-②得Sn-S(n-1)=2an+3n-7-[2a(n-1)+3(
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-
因为6Sn=(an+1)(an+2)(1)所以6Sn-1=(an-1+1)(an-1+2)(2)(1)-(2)则an-an-1=3所以an是等差数列因为6Sn=(an+1)(an+2)可知S1=a1=
(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=
已知Sn=2An-1取n=1得:S1=2A1-1又因为S1=A1,解上述方程可得:A1=1Sn=2An-1S(n-1)=2A(n-1)-1注:"n-1"为下标上下两式相减得:Sn-S(n-1)=2An