已知数列满足an 1=2an an-1,
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/22 14:45:47
1=√a1a2=√2b2=b1q=√a2a3,a3=b1^2q^2/a2=q^2bn=b1q^(n-1)=√anan+1bn+2=b1q^(n+1)=√an+1an+2anan+1=2q^(n-1)a
anan+1-2an=0anan+1=2anan+1=2所以a2=2a3=2a4=2
an*a(n-1)=a(n-1)+(-1)^nan=1+(-1)^n/a(n-1)a1=1a2=1+1/1=2a3=1-1/2=1/2a4=1+1/1/2=3a5=1-1/3=2/3a5/a3=(2/
(1)证:ana(n-1)=2a(n-1)-1a(n-1)=0时,0=-1,等式恒不成立,因此数列各项均不为0a(n-1)=1/2时,ana(n-1)=0,an和a(n-1)中至少有1个为0,与各项不
An+1=an/1+2an两边去倒数1/an+1-1/an=21/an=1+(n+1)*2=2n+3an=1/[2n+3]a1a2+a2a3+……+anan+1=1/2[1/a1-1/a2+1/a2-
∵数列{a[n]}满足4a[n+1]-a[n]a[n+1]+2a[n]=9∴(4-a[n])a[n+1]=9-2a[n]即:a[n+1]=(2a[n]-9)/(a[n]-4)∵a[1]=1∴a[2]=
解:an*a(n+1)+a(n+1)=2an两边同时除以an*(an+1)得:1+1/an=2/a(n+1)设:bn=1/an则:2b(n+1)=bn+12[b(n+1)-1]=bn-1[b(n+1)
1=√a1a2=√2b2=b1q=√a2a3,a3=b1^2q^2/a2=q^2bn=b1q^(n-1)=√anan+1bn+2=b1q^(n+1)=√an+1an+2anan+1=2q^(n-1)a
AnAn-1=An-1-AnAnAn-1+An=An-1An=An-1/(A(n-1)+1)n>=2A1=1/3A2=A1/(A1+1)=1/3/(1/3+1)=1/4A3=A2/(A2+1)=A1/
已知数列An满足:A1=1,A2=a(a>0),数列Bn=AnAn+1(1)若AN是等差数列,且B3=12,求a的值及AN通项共识你看看那B3=12应该=A3*A3+1(这就是利用Bn=AnAn+1)
(Ⅰ)由bn=an-1得an=bn+1代入2an=1+anan+1得2(bn+1)=1+(bn+1)(bn+1+1)整理得bnbn+1+bn+1-bn=0从而有1bn+1−1bn=1∴b1=a1-1=
(1)∵{an}是等差数列,a1=1,a2=a(a>0),∴an=1+(n-1)(a-1).又b3=45,∴a3a5=45,即(2a-1)(4a-3)=45,解得a=2或a=-74(舍去),…(5分)
由(an-1-an)/(anan-1)=(an-an+1)/(anan+1)(n≥2),得到1/an-1/a(n-1)=1/a(n+1)-1/an{1/an}是等差数列,而且公差d=1/a2-1/a1
(1)b1=√2,bn=√2*q^(n-1)(bn+1/bn)^2=an+2/an=q^2(2)Cn+1=a2n+1+2a2n+2=q*a2n-1+2q*a2n=q*(a2n-1+2a2n)=q*Cn
麻烦你把你的问题写清楚了,那些an-1到底都是第N-1项还是第N项再-1.如果是第N-1项你可以这样写a(n-1).再问:an*a(n-1)+1=2*a(n-1)bn=1/a(n-1)
anan-1=an-1-an两边同除anan-11=1/an-1/a(n-1)∴1/an=1/a1+(n-1)×1=2+n-1=n+1∴an=1/(n+1),n≥2n=1时,a1=1/(1+1)=1/
√[a(n-1)]-√[an]=√[ana(n-1)]两边同时除以√[ana(n-1)]得:1/√[an]-1/√[an(n-1)]=1令bn=1/√[an]则bn-b(n-1)=1,b1=1∴bn是
∵1=2,an+1=1+an1−an(n∈N*),∴a2=1+a11−a1=1+21−2=-3,a3=1+a21−a2=1−31+3=−12a4=1+a31−a3=1−121+12=13a5=1+a4
(1)∵anan+1=2n,∴anan-1=2n-1,两式相比:an+1an−1=2,∴数列{an}的奇数项成等比数列,偶数项成等比数列,∵a1=1,a nan+1=2n(n∈N*)∴a1=