已知m是等差数列别人是等比数列量大于零恒成立若a2等于b2且a8等于8折
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/13 19:52:38
(1)由题意可知,2a3=a1+a2,即2aq2-q-1=0,∴q=1或q=-12;(II)q=1时,Sn=2n+n(n−1)2=n(n+3)2,∵n≥2,∴Sn-bn=Sn-1=(n−1)(n+2)
lgA(n+1)-lgAn=q(q为常数)lgA(n+1)/An=dqA(n+1)/An=10^q所以{An}是等比数列
取数列{lgan}中的任意两项lgan和lga(n-1),那么必定有lgan-lga(n-1)=k=常数所以有lg[an/a(n-1)]=k那么an/a(n-1)=e^k所以数列{an}中的任意两项a
∵{An}是等差数列∴An-A(n-1)=d(d为公差)∵Bn=kAn+m∴B(n-1)=kA(n-1)+m∴Bn-B(n-1)=kAn+m-[kA(n-1)+m]=k[An-A(n-1)]=kd这个
(1)根据题意,设公差为d则a3=a1+2d=2d+1a9=a1+8d=8d+1有(2d+1)^2=8d+1d=1故通项:an=n(2)根据题意,设公比为q则b2=qb3=q^2有q-0.5q^2=0
an^bn/an^b(n-1)=an^[bn-b(n-1)]=an^d,这是个常数,所以是等比数列bn-b(n-1)=d再问:d是什么再答:公差啦,高二数学书丽有的再答:采纳我吧,3q了
(1)设公比为q∵a7=1∴a4=1/q³,a5=1/q²,a6=1/q∵a4,a5+1,a6成等差数列∴a4+a6=2(a5+1)即1/q³+1/q=2(1/q
6m+7=3k+16(m+1)=3kk=2m+2q=bn/bn-1=an+1/an-1an+1-(an-1)=2d两个联立an-1=1+2d/q是常数所以an是常数列bn也是常数列,且bn=1
a1*p=a2a1*p^3=a4,a1*p-a1=a1*p^3-a1*Pp-1=p^(p^2-1);(p-1)(p*(p+1)-1)=0,p=1,或p^2+p-1=0,p=(-1+√5)/2,p=(-
a1,a2,a4成等差数列2a2=a1+a4即2a1*q=a1+a1q^3a1不为0所以:2q=1+q^3q^3-2q+1=0q^3-q^2+q^2-2q+1=0q^2*(q-1)+(q-1)^2=0
a1,a2,a4成等差数列所以2a2=a1+a4{an}是等比数列a2=a1qa4=a1q^3所以2×a1q=a1+a1q^3即:q^3-2q+1=0(q-1)(q^2+q-1)=0q=1或q=(-1
a1,a2,a4成等差数列所以2a2=a1+a4{an}是等比数列a2=a1qa4=a1q^3所以2×a1q=a1+a1q^3即:q^3-2q+1=0(q-1)(q^2+q-1)=0q=1或q=(-1
a2=a1qa8=a1q^7a5=a1q^42a8=a2+a52a1q^7=a1q+a1q^42q^6=1+q^32q^6=1+q^32q^6-q^3-1=0(2q^3+1)(q^3-1)=0q^3=
a1+a2=a3=b2+b3有问题,是不是a1+a2+a3=b2+b3
∵an=a1q(n-1),bn=b1+(n-1)d,∵a6=b7∴a1q5=b1+6da3+a9=a1q2+a1q8b4+b10=2(b1+6d)=2b7=2a6a3+a9-2a6=a1q2+a1q8
(1).由a(m)+a(m+1)=a(k)知道3m+3(m+1)+1=3k+1,整理后有k-2m=4/3,而m,k均是N+,则k-2m也是整数,故而不存在m,k∈N+,使a(m)+a(m+1)=a(k
(1)将a4+a4q^2=2*(a4q+1)与a4q^3=1联立,得q=1/2,a4=8,所以an=64q^(n-1)(n>=1,n∈R+)(2)Sn=64[1-(1/2)^n]/(1-1/2)=12
Sn=a1*(q^n)/(1-q)Sm+S2m=2S3m,即q^m+q^2m=2q^3m得{1+q^m=2q^2m.}am=a1*q^(m-1),a2m=a1*q^(2m-1),a3m=a1*q^(3
设a,x,y,b依次成等差数列的公差为d,则:x=a+d,y=a+2d,b=a+3d;a,m,n,b依次成等比数列的公比为q,则:m=aq,n=aq2,b=aq3,所以有a+3d=aq3得到3d=aq
设此连续三项为al,am,an,因为他们成等比数列,故公比q=am/al=an/am=(-an)/(-am),由等比定理得q=(am-an)/(al-am)=(m-n)d/(l-m)d=(n-m)/(