已知数列an an不等于0 ansn 1-an 1sn
来源:学生作业帮助网 编辑:作业帮 时间:2024/05/14 04:00:50
点一下,Bn=√(AnAn+1),则1/An=A(n-1)/(Bn)^2,A(n-1)=(B(n-2))^2/A(n-2),1/A(n-2)=A(n-3)/(B(n-3))^2...1/An=A(n-
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
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
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
1.d=A2-A1=a-1An=A1+(n-1)d=1+(n-1)(a-1)A3=2a-1,A4=3a-2,B3=A3A4=(2a-1)*(3a-2)=126a^2-7a+2=12(6a+5)*(a-
设A1A2=a则:由于在数列{An}中An小于0故a>0,且An+1An+2/AnAn+1>0即q>0;由题中:2AnAn+1+An+1An+2>An+2An+3得2aq^(n-1)+aq^n>aq^
根据bn=1/(an*a(n+1)),我们知道,bn=[1/an-1/a(n+1)]/d.因此,Sn=[1/a1-1/a2+1/a2-1/a3+...+1/a(n-1)-1/an]/d=[1/a1-1
1)b3=(a3)^2+1a3=1+2dd=a-1所以12=(1+2a-2)^2+1a=(√11+1)/2an=1+(n-1)*(√11-1)/22)an=a^(n-1)bn=a^[2(n-1)]+1
∵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
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
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:
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)
1.由题意可知a1a2=1*2=2ana(n+1)=2*q^(n-1)ana(n+1)+a(n+1)a(n+2)>a(n+2)a(n+3)即2*q^(n-1)+2*q^n>2*q^(n+1)两边同时乘
【解】(1)方程A(k)(X^2)+2A(k+1)X+A(k+2)=0,则其Δ=4[A(k+1)^2-A(k)*A(k+2)]=4[[A(k)+d]^2-A(k)*[A(k)+2d]]=4d^2>0;
(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)
(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
(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=