已知数列an an=2n-1,n为偶数 3n,n为奇数
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数列an中,a1=3,a=2an-1,∴a-1=2(an-1),∴an-1=(a1-1)*2^(n-1)=2^n,∴an=2^n+1,∴bn=2n/{(2^n+1)(2^(n+1)+1]},∴Sn=2
在原式基础上,再写一相同结构等式,到an+2结束.减去原式便得到:1/(an+1)an=n+1/(an+1)(an+2)-n/anan+1整理得…你题目可能出错了,不是等差数列.我们假设公差为d.那么
n=1,S1=a1=2,n>1,an=Sn-S(n-1)=2n,n=1时也适合,故:an=2nbn=(1/4)·1/n(n+1)4bn=1/n(n+1)=1/n-1/(n+1),所以:4Tn=[(1-
(Ⅰ)∵数列{an}满足a1=1,an+1=2n+1anan+2n(n∈N*),∴2n+1an+1=2nan+1,即2n+1an+1−2nan=1,∴数列{2nan}是公差为1的等差数列.(Ⅱ)由(Ⅰ
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/
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-
1/a(n+1)=an+2/2an=1/2+1/an所以,{1/an}是公差为1/2的等差数列1/an=1/a1+(n-1)*1/2=(n+1)/2an=2/(n+1)a(n+1)=2/(n+3)an
an*a(n+1)=2^na(n-1)*an=2^(n-1)所以:a(n+1)/a(n-1)=2a1=1,所以a2=2(此时分奇数和偶数讨论)a(2n+1)=2^n,a(2n)=2^n所以a9=2^4
an=1/(n+1)+2/(n+1)+...+n/(n+1)=(1+2+...+n)/(n+1)=[n(n+1)/2]/(n+1)=n/2bn=2/[ana(n+1)]=2[(n/2)(n+1)/2]
a1=S1=20-1=19,an=Sn-Sn-1=-2n+21,n≥2a1时也符合∴an=-2n+21anan+1=(-2n+21)(-2n+19)<0∴192<n<212∵n∈N∴n=10故答案为:
Sn=2n^2+nSn-1=2(n-1)^2+n-1an=Sn-Sn-1=4n-1lim[1/a1a2+1/a2a3+1/a3a4+...+1/anan+1]=lim[1/3*1/7+1/7*1/11
(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
∵a1=3,an-anan+1=1(n∈N+),∴3-3a2=1,∴a2=23,23−23a3=1,∴a3=−12,−12+12a4=1,a4=3,∴数列{an}是周期为3的数列,且a1•a2•a3=
参考百度,】an=2n,即246810121416a1a2+…+anan+1=An,即8244880120168……An=4n(n+1)平方和的公式为S=n(n+1)(2n+1)/6所以,Sn=4×n
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)
由题意得1/a1a2+1/a2a3…1/anan-1=(n-1)/a1an①原式-①得1/anan+1=n/a1an+1-(n-1)a1an整理得2=nan-(n-1)an+1两边同时除以n(n-1)
由题意:n=1时,a2*a1=a2*1=2,即a2=2n=2时,a2*a3=4,即a3=2当n>=2时,anan+1=2^nan-1an=2^(n-1)故an+1/an-1=2所以隔项成等比数列当n为
(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=