设正数等比数列中,A2=4,a4=16,求limlgan=1

设公比为q,公差为d,s=a1^50*q^（1+2+……+49）=a1^50*q^1225,t=a1^50*q^[(n-50)+(n-49)+……+(n-1)]=a1^50*q^[25(2n-51)]

log2a1+log2a2+……+log2a8=log2(a1×a2×…×a8)∵等比数列∴a1a8=a2a7=a3a6=a4a5=32∴log2(a1×a2×…×a8)=log2(32^4)=log

1.a1+a2+a3=6a2+a3+a4=q*a1+q*a2+q*a3=q(a1+a2+a3)=6q=-3q=-1/2a1+a2+a3=a1+q*a1+q²*a1=a1-a1/2+a1/4=

a1+a2=1,a3+a4=4,则a5+a6=因为是等比数列,设公比为n,则有,a2=na1,a3=n²a1,a4=n³a1..然后把那两个式子里面的n²提出来,得出n&

31

公比为q,a1=a2/q,a3=a2qa1×a2×a3=a2³同理,a4×a5×a6=a5³...a28×a29×a30=a29³因此a1×a2×a3×...×a30=(

因为a2+a3+a4=15,所以3a1+6d=15即a1+2d=a3=5又a2,a3-1,a4成等比数列即5-d,4,5+d成等比数列于是有(5-d)(5+d)=16解得d=3（因为公差d为正数,所以

（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）∵

设公比为q,则q>0a3=a2+4a1q^2=a1q+4a1=2代入,整理,得q^2-q-2=0(q+1)(q-2)=0q=-1(舍去)或q=2Sn=a1(q^n-1)/(q-1)=2×(2^n-1)

在等差数列{an}中,a2+a4=2a3,代入a2+a3+a4=15,得a3=5,∴a2+a4=10,又a2,a3-1,a4成等比数列,∴(a3-1)²=a2a4,即a2a4=16,∵公差d

∵a2*a4=4∴a3=2.q=1/2.an=2^(4-n)2^(9-3n)>1/9.9-3n>=-3n

设公比为q，则∵各项都是正数的等比数列{an}中，3a1，12a3，2a2成等差数列，∴a3=3a1+2a2，∴q2=3+2q，∵q＞0，∴q=3，∴a2012+a2014a2013+a2011=a2

a1+a2=a1+a1q=a1(1+q)=21a3+a4=a1q^2+a1q^3=a1q^2(1+q)=822式除1式得q^2=4q=±2分别代入1式得a1=2/3a1=-2(舍去q=-2)S8=a1

a1(q+q^3)=4a1(1+q+q^2)=14两式相除：(q+q^3)/(1+q+q^2)=2/7求得qan+an+1+an+2=（a1+a2+a3）*q^(n-1)>1/9关键是求q说实在的,我

a3*a8=81a1*a10=a2*a9=a3*a8=……=a5*a6=81log3(a1)+log3(a2)+…+log3(a10)=log3(a1*a2*……*a10)=log3[(a1*a10)

等比数列,则：a1a3=(a2)²,a3a5=(a4)²,则：a1a3＋2a2a4＋a3a5=(a2)²＋2a2a4＋(a4)²=(a2＋a4)²=1

A1*A2*A3*……A30=(A15*A16)^15=2^15A15*A16=2A3*A6*A9*……A30=(A15*A18)^5=(A15*A16*2)^5=(2*2)^5=2^10

证明：（1）左边=log2a+b+ca+log2a+b−cb＝log2(a+b+ca•a+b−cb)=log2(a+b)2−c2ab＝log2a2+2ab+b2−c2ab＝log22ab+c2−c2a

答案：3/2lg2由a2=4,a4=16,求得a1=2,q=2,即an=2×2^(n-1)=2^n所以(lga(n+1)+lga(n+2)+...+lga(2n))/n^2=lg2((n+1)+(n+

因为an>0,a2=4,a4=16所以q=2,a1=2所以lim(lgan+1+lgan+2+...+lga2n)/(n^2)=lim(n/2*lg(an+1*a2n))/(n^2)=lim(lg(a