设a1=a2 an 1=an an-1
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/05 03:11:13
![设a1=a2 an 1=an an-1](/uploads/image/f/7245205-61-5.jpg?t=%E8%AE%BEa1%3Da2+an+1%3Dan+an-1)
an*an+1/an-1*an=q→an+1/an-1=q→an+2/an=q→a(2k+1)=a(2k-1)*q;a(2k)=a(2k-2)*qa1=1,a2=r→bn=q^(n-1)+r*q^(n
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
如果an不等于0有a(n+1)/an=2-a(n-1)a1=1,有a3=a2=1由数学归纳法可知an=1是常数列再问:不好意思是an+1+a(n+1)an-2an=0a1=1求通项再答:。。。。这个简
anan+1-2an=0anan+1=2anan+1=2所以a2=2a3=2a4=2
1/a1a2+1/a2a3+…+1/anan+1=[(a2-a1)/a1a2+(a3-a2)/a2a3+…+(a(n+1)-a(n))/anan+1]/d=[1/a1-1/a2+1/a2-1/a3+.
∵a1,a3,a5,a7成公比为q的等比数列∴a1,a1q,a1q²,a1q³∵a2,a4,a6成功差为1的等差数列∴a2,a2+d,a2+2d即a2,a2+1,a2+2∵1=a1
1/a1=1d=2所以1/an=(2n-1)所以原式=1/1*3+1/3*5+……+1/(2n-1)(2n+1)=(1/2)(1-1/3)+(1/2)(1/3-1/5)+……+(1/2)[1/(2n-
∵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)
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
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)∵数列{a[n]}满足条件:a[1]=1,a[2]=r,且数列{a[n]a[n+1]}是公比为q的等比数列∴q≠0,r≠0,且a[n]a[n+1]=a[1]a[2]q^(n-1)=rq^(n-1
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
∵数列{a[n]}满足条件:a[1]=1,a[2]=r,且数列{a[n]a[n+1]}是公比为q的等比数列∴q≠0,r≠0,且a[n]a[n+1]=a[1]a[2]q^(n-1)=rq^(n-1)∵a
1、a1+a4=a2-d+a3+d=a2+a3=8,又有a2a3=15易求出:a2=3,a3=5,公差d=2因此,a1=1,通项公式an=1+2(n-1)=2n+12、bn=1/anan=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
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=
把3anan-1+an-an-1=0两边同时除以anan-1,变为1/an-1/an-1=3,{1/an}为等差数列,求出an=1/(3n-2)“{bn}满足bn=1、an”这句话之后看不懂你的意思.