设an是等比数列,公比Q等于根号2Asn为an的前n项和.
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![设an是等比数列,公比Q等于根号2Asn为an的前n项和.](/uploads/image/f/7246185-33-5.jpg?t=%E8%AE%BEan%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E5%85%AC%E6%AF%94Q%E7%AD%89%E4%BA%8E%E6%A0%B9%E5%8F%B72Asn%E4%B8%BAan%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C.)
qSn+a1=Sn+a(n+1)Sn=a1(1-q^n)/1-q=a1(2^n-1)S4/a2=a1*(2^4-1)/2a1=15/2
q=2等比数列a2=2a1s4=[a1/(1-2)]*(1-2&4)=(-a1)*(-15)=15a1s4/a2=15a1/2a1=15/2
S4=a1+a2+a3+a4S2=a1+a2所以S4/S2=1+q^2=1+4=5
S4/a2=(15a1)/(2a1)=15/2ps.高考过后我的智商呈直线下降...我胆敢回答你的问题也是因为我高三拿过数学奥赛全国三等奖,可能还残余点东西吧.错了我就要说声对不起了
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
S4=A2/2+A2+A2*2+A2*2*2=15/2*A2所以S4/A2=15/2
loganan+1-log(an-1)an=logan(an×q)-log(an-1)(an-1×q)=1+loganq-1-log(an-1)q=loganq-log(an-1)q<0所以递减
{1+an}的首项为3(1+an)=3*2^(n-1)1+a(6)=3*2^5=96a(6)=95
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)
∵-a5,a4,a6成等差数列,∴-a5+a6=2a4,∴-a4q+a4q2=2a4,∴q2-q-2=0,∴(q+1)(q-2)=0,∴q=-1或2.故选:C.
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
∵{an}是公比q≠1的等比数列,且a2=9,a3+a4=18,∴a1q=9a1q2+a1q3=18,整理,得q2+q-2=0,解得q=-2,或q=1(舍)故选A.
4x²-8x+3=0(2x-3)(2x-1)=0x1=1/2,x2=3/2于是a2004=1/2,a2005=3/2,q=3a2006=a2004*q²,a2007=a2005*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
-1或者2再问:答案对了,求过程讲解,十分感谢再答:再问:谢啦再答:不客气
a5=a4*qa7=a4*q^3a6=a4*q^22(a5+a7)=a4+a62(a4*q+a4*q^3)=a4+a4*q^2a4不等于0两边同时÷a42q+2q^3=1+q^22q(1+q^2)=1
(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)