等比数列q等于0
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![等比数列q等于0](/uploads/image/f/6646600-64-0.jpg?t=%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97q%E7%AD%89%E4%BA%8E0)
解题思路:根据题目条件,由等比数列的知识可求解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/inc
设an=a+(n-1)d那么a3=a+2d=10a7=a+6da10=a+9d由于等比,a7为比例中项,a7²=a3*a10(a+6d)²=(a+2d)(a+9d)da+18d&s
已知等比数列{an}的公比q>1,a17^2=a24,求使a1+a2+a3+……+an>1/a1+1/a2+1/a3+……+1/an成立的n的取值范围.【解】a17^2=a24,a1^2q^32=a1
⑴若q=1,显然所求极限为na1/(n-5)a1=n/(n-5)的极限,易知极限是1q≠1时,所求的实际是Sn/(Sn-S5)的极限Sn/(Sn-S5)=a1(q^n-1)/[a1(q^n-1)/(q
因为a2+a5=9/4,a3.a4=1/2所以a2(1+q^3)=9/4,a2^2.q^3=1/2(计算过程把q^3看作整体来解)即a2=2,q=1/2所以an=4.(1/2)^(n-1)
∵{an}是等比数列,∴an+2=an+1+2an,可化为a1qn+1=a1qn+2a1qn-1,∴q2-q-2=0.∵q<0,∴q=-1.∵a2=a1q=1,∴a1=-1.∴数列{an}的前2010
∵-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.
这个图片不知道行不行啊再问:{an+1}为等比数列怎麽会有An+1+An-1=An再答:这是按照上面的公式算出来的啊,是等于2An因为an是等比数列,所以an+1*an-1=an*an
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
因为a2、a3、a6成等比数列,所以a32=a2•a6⇒(a1+2d)2=(a1+d)(a1+5d)⇒2a1d+d2=0.∵d≠0,∴d=-2a1.∴q=a3a2=a1+2da1+d=3.故选C.
设an=a1×q^(n-1)an+2=an+a(n+1)a1×q^(n+1)=a1×q^(n-1)+a1×q^nq^2=1+qq=(1±√5)/2再问:q^2=1+q这部是什么意思再答:a1×q^(n
设三个数为a/qaaqa/q*a*aq=a^3=27a^2/q^2+a^2+a^2q^2=91所以a=39/q^2+9+9q^2=91设q^2=t9/t+9t-82=09t^2-82t+9=0t=1/
a2+a3=a1q+a1q²=6a1=1所以q²+q-6=0(q-2)(q+3)=0q>1q=2所以an=2的(n-1)次方
a2=a1×q^(5-1)=2×(1/2)^4=2×1/16=1/8
-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
s5/s2=-11,答案对吗?采纳哦
a1+a1q+a1q^2=3a1+a1q+a1q^2+a1q^3+a1q^4+a1q^5=27a1q^3+a1q^4+a1q^5=24q^3(q1+a1q+a1q^2)=3q^3=24q^2=8,q=