设等比数列的公比S10=8
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S4=a1+a2+a3+a4=a2/q+a2+a2*q+a2*q^2S4/a2=1/q+1+q+q^2=7.5
第二题:1/(X-1)=1X>=2所以不等式解集为X=2第一题公比q若为正数的话,哪么应该大于1,因为要是q
S4=a1(1-q^4)/(1-q)=5a1(1-q^2)/(1-q)1+q^2=5q^2=4因为q
(1)因为Sn=a1(1-q^n)/(1-q)所以S10/S5=(1-q^10)/(1-q^5)=31/32又因为(1-q^10)/(1-q^5)=(1-q^5)(1+q^5)/(1-q^5)=1+q
S10=a1(1-q^n)/(1-q)=a1(1-q^10)/(1-q)=8S20/(1+q^10)=a1(1-q^20)/(1-q)/(1+q^10)=a1(1+q^10)(1-q^10)/(1-q
q≠1,不然的话,S10:S5=2:1≠1:2.S10=a1(1-q^10)/(1-q),S5=a1(1-q^5)/(1-q),S10:S5=1+q^5=1/2,q^5=-1/2.S15:S5=1+q
{1+an}的首项为3(1+an)=3*2^(n-1)1+a(6)=3*2^5=96a(6)=95
∵{an+c}是等比数列∴(a1+c)(a3+c)=(a2+c)2即a1a3+c(a1+a3)+c2=a22+2a2c+c2∵a1a3=a22∴(a1+a3)c=2a2c即a1c(1+q2)=2a1q
首先得求的a1a4=5s2...a1q^3=5(a1+a1q)又.a3=a1q^2=2...所以.2q=5(a1+a1q)得.a1=(2q)/(5(1+q))又因为.a3=a1q^2=2得.q=1.2
等比数列an=a1*q^(n-1),Sn=a1(1-q^n)/(1-q)∴a3=2=a1*q^(3-1)=a1*q^2S4=5S2=>a1(1-q^4)/(1-q)=5*a1(1-q^2)/(1-q)
Sn=a1*(1-q^n)/(1+q)S20/(1+q^10)=a1*(1-q^20)/((1+q)(1+q^10))=a1*(1-q^10)(1+q^10)/((1+q)(1+q^10))=a1*(
S20/(1+q^10)=[a1(1-q^20)/(1-q)]/(1+q^10)=a1(1-q^20)/[(1-q)(1+q^10)]=a1(1-q^10)(1+q^10)/[(1-q)(1+q^10
(Ⅰ)∵设{an}是公比为正数的等比数列∴设其公比为q,q>0∵a3=a2+4,a1=2∴2×q2=2×q+4解得q=2或q=-1∵q>0∴q=2∴{an}的通项公式为an=2×2n-1=2n(Ⅱ)∵
S4=a1(1-q4)/(1-q),S2=a1(1-q2)/(1-q),已知S4=5S2,则a1(1-q4)/(1-q)=5a1(1-q2)/(1-q),即q=±2,又公比q
a2004和a2005是方程4x平方+8x+3=0的两根4x^2+8x+3=0(2x+3)(2x+1)=0x=-3/2x=-1/2∵q>1∴a2004=-1/2a2005=-3/2q=a2005/a2
(1)∵(a6+a7+a8+a9+a10)/(a1+a2+a3+a4+a5)=(a1*q^5+a2*q^5+a3*q^5+a4*q^5+a5*q^5)/(a1+a2+a3+a4+a5)=q^5∴S10
(1)首项是a1,公比是qS5+S10=2S15(S中都乘有一项a1/1-q,由于等式两边都有在此略去!)(1-q^5)+(1-q^10)=2(1-q^15)q^5(2q^10-q^5+1)=0∵q≠
S4=a1(1-q4)/(1-q),S2=a1(1-q2)/(1-q),已知S4=5S2,则a1(1-q4)/(1-q)=5a1(1-q2)/(1-q),即q=±2,又公比q
证明:假设{Cn}为公比为q的等比数列设{an}的公比为q1,{bn}的公比为q2,则Cn=C1*q^(n-1)而C1=a1+b1,故Cn=a1*q^(n-1)+b1*q^(n-1)又因为an=a1*
S10=a1+a2+.+a10=m;S20=a1+a2+,+a10+a11+a12+...+a20=m+q^10×m=m×(1+q^10);S30=a1+a2+...+a10+a11+a12+...+