已知数列{an}与{bn}满足a(n 1)-qb(n 1)
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![已知数列{an}与{bn}满足a(n 1)-qb(n 1)](/uploads/image/f/4267520-8-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%8E%7Bbn%7D%E6%BB%A1%E8%B6%B3a%28n+1%29-qb%28n+1%29)
(n+1)/bn=2∴bn=b1×2^(n-1)b1=a2-a1=3-1=2∴bn=2^n∴a(n+1)-an=2^n∴a2-a1=2a3-a2=2^2a4-a3=2^3……an-a(n-1)=2^(
即对任意n∈N,(a+n)/(a+n-1)≥(a+8)/(a+7)两边同减1:1/(a+n-1)≥1/(a+7)此不等式可分三种情况:(1)a+7≥a+n-1〉0显然n≥8时不成立(2)0〉a+n-1
(1)∵等比数列{bn}的公比为3∴bn+1bn=3an+13an=3an+1−an=3∴an+1-an=1∴{an}是等差数列(2)∵a1=1,an+1-an=1∴an=n则cn=1anan+1=1
这个题目可用倒推法解.首先,要使BN>1,那么LOG(AN)>1/2,换句话说也就是AN>10根据AN=-2N+24得,N
n=1-an,第二个式子代入bn=1-anbn+1=(1-an)/(1-an^2)=1/(1+an)an+1=1-bn+1=an/(1+an)求倒数1/(an+1)=1+1/an令cn=1/an,cn
(1){an}是等差数列,a1=1,a2=a(a>0),an=1+(n-1)(a-1)a3=2a-1,a4=3a-2b3=a3*a4=(2a-1)(3a-2)=12a=2,或-5/6(舍去)所以a=2
1.bn/b(n-1)=3[an-a(n-1)]=q所以an-a(n-1)=log(3)q2.a2=13a8=1d=-2an=17-2n3.n8Tn=-[a1+.an]+2[a1+.+a8=n^2-1
解(1)证明:由bn=an3n,得bn+1=an+13n+1,∴bn+1−bn=an+13n+1−an3n=13---------------------(2分)所以数列{bn}是等差数列,首项b1=
1.证明:因为bn,a(n+1),b(n+1)成等比数列,所以[a(n+1)]²=bnxb(n+1)(n∈N*)a(n+1)=√[bnxb(n+1)]所以an=√[bnxb(n-1)](n≥
设{bn}共比为q则q=b(n+1)/b(n)=3^a(n+1)/3^a(n)=3^[a(n+1)-a(n)]所以a(n+1)-a(n)=log(3,q)是定值,所以{an}是等差数列若a8=a13=
根据提示算出A3=-18/7;A4=-32/7;A5=22/7;A6=36/7;A7=-26/7;A8=-40/7;C1=A1+A3=2+(-18/7)=-4/7C2=A3+A5=-18/7+22/7
设an=a1*2^(n-1)b(n+1)=an+bn故有:b(n+1)-bn=an=a1*2^(n-1)bn-b(n-1)=a1*2^(n-1)b(n-1)-b(n-2)=a1*2^(n-2)…………
由条件知:tbn+1=2b(n+1),且t≠2.可得b(n+1)+1/(t-2)=(t/2)[bn+1/(t-2)].由f(b)≠g(b),t≠2,t≠0,可知b+1/(t-2)≠0,t/2≠0,所以
d(n)=2^n+n,p(1)=d(1)=2^1+1=3,p(n+1)=d(n+1)+d(n)=2^(n+1)+(n+1)+2^n+n=3*2^n+2n+1,L(2n-1)=d(2n-1)=2^(2n
n=b1.q^(n-1)bn=an-3nan=bn+3n=b1.q^(n-1)+3nSn=a1+a2+...+an=b1(q^n-1)/(q-1)+3n(n+1)/2
(1)证明:由bn=3-nan得an=3nbn,则an+1=3n+1bn+1.代入an+1-3an=3n中,得3n+1bn+1-3n+1bn=3n,即得bn+1-bn=13.所以数列{bn}是等差数列
1、证明:a1=λ,a2=(2/3)a1+1-4=2λ/3-3,a3=(2/3)a2+2-4=4λ/9-4.若λ=0,a1=0,显然{an}不是等比数列;若λ≠0,则a2/a1=2/3-3/λ,a3/
1.bn=a1+a2+a3...an\nnbn=a1+a2+a3...an=n^3an=n^3-(n-1)^3=3n^2-3n+12.令a1+a2+a3...an=Snbn=b+(n-1)dbn=a1
(1)a(n+1)-an=(n+1+2013)-(n+2013)=1∴b(n+1)-bn=cn/[a(n+1)-an]=cn=2^n+n∴bn-b(n-1)=2^(n-1)+n-1...b2-b1=2
a(n+1)+b(n+1)=1,b(n+1)=(1-an)/(1-an²)=1/(1+an),a(n+1)+1/(1+an)=1,a(n+1)an+a(n+1)+1=1+an,a(n+1)a