若正项等比数列{An}的公比q不等于1,且a3,a5,a6

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

（1）由a3=14=a1q2，以及q=-12可得a1=1．∴数列{an}的前n项和Sn=1×[1−(−12)n]1+12=2−2•(−12)n3．（2）证明：对任意k∈N+，2ak+2-（ak+ak+

我猜你的题目给出的条件是a(n+2)=a(n+1)+2an,就像楼上所列正解如下a3=a2+2a1=2a1+1a4=a3+2a2=2a1+1+2=2a1+3又an为等比数列,a2=a1*q,a3=a1

因为am,an,ap成等比数列,则由等比中项,有：（an)^2=am*ap(a1*q^(n-1))^2=a1*q^(m-1)*a1*q^(p-1)(这是把通项公式代入）则消去a1,（q^(n-1))^

S4=a1(1-q^4)/(1-q)=5a1(1-q^2)/(1-q)1+q^2=5q^2=4因为q

S1=a1(1-q)/(1-q),S2=a1(1-q^2)/(1-q),...,Sn=a1(1-q^n)/(1-q).S1+S2+...+Sn=[a1/(1-q)]*[1-q+1-q^2+...+1-

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

a2=qa1,a3=q^2a1,且a1.a3.a2成等差数列,则2a3=a1+a22q^2a1=a1+qa1,即2q^2=1+q,解得：q=1或q=-1/2

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)

(1)a3*a4=a2*a5=1/2a2+a5=9/4-1

∵{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)

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)S1→3=a1(1+q+q^2)=a1*(1-q^3)/(1-q)S4→6=a4(1+q+q^2)=a1*(1-q^3)/(1-q)*q^3S7→9=a7(1+q+q^2)=a1*(1-q^3)

a1(1+q)=1,a1q^2(1+q)=4q^2=4,q=-2a4+a5=a1q^3(1+q)=(a3+a4)*q=-8

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

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

2a4=-a5+a62a4=-a4q+a4q^22a4=-a4q+a4q^2a4q^2-a4q-2a4=0a4(q^2-q-2)=0a4(q-2)(q+1)=0(q-2)(q+1)=0q=2或q=-1

∵等比数列{an}中，公比q＝12，且log2a1+log2a2+…+log2a10=55=log2（a1a2…a10）=log2 (a1a10) 5，∴(a1a10)5=255，