{an}各项为整数,an² 2an=4Sn
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(1)由已知得:a2=2a1,a3=4a1,所以S3/a3=(a1+a2+a3)/a3=(a1+2a1+4a1)/(4a1)=7/4(2)这一步不知道要求什么结果,我就帮你求出an吧!由a5a6=81
设原来公比是q√an存在则q>0a(n+1)/an=q则√a(n+1)/√an=√q,所以是等比数列
再问:……看不清楚……再答:你的放大不了?(I)由a1=S1=-(a1+1)(a1+2),解得a1=1或a1=2,由假设a1=S1>1,因此a1=2,又由an+1=Sn+1-Sn=-(an+1+1)(
Sn=1/2(an+1/an),an>0令x=1得:S1=1/2(a1+1/a1)解得a1=1注意到an=Sn-S(n-1),上式可化为:Sn=1/2(Sn-S(n-1)+1/(Sn-S(n-1)))
那么我把Aˇ〔3/2〕n+1理解成A[n+1]的3/2次方了递推式可以化成A[n]/A[n+1]^2=(A[n+1]/A[n+2]^2)^(-1/2)两边取对数得到log(A[n]/A[n+1]^2)
设等差数列的公差为d,则a3=a5-2d=6-2d,an1=a5+(n1-5)d=6+(n1-5)d.∵a3,a5,an1成等比数列,∴a52=a3an1化简即(6n1-42)d-2(n1-5)d2=
S1=a1所以a1=3a1-2a1=1S2=3a2-2所以a1+a2=3a2-22a2=3a2=3/2和各项都是整数矛盾无解
设bn=根号an所以A(n-1)-An=(2倍根号An)+1等于根号[b(n-1)]^2-bn^2=2bn+1即[b(n-1)]^2=(bn+1)^2因为{a}中各项为正数,且a1=2所以b(n-1)
∵a2*a4=4∴a3=2.q=1/2.an=2^(4-n)2^(9-3n)>1/9.9-3n>=-3n
∵(an+1)²-an+1×an-2an²=0∴(an+1+an)(an+1-2an)=0∴an+1-2an=0,an+1+an=0(舍去)∴an+1=2an∴an是等比数列,设a
a1(q+q^3)=4a1(1+q+q^2)=14两式相除:(q+q^3)/(1+q+q^2)=2/7求得qan+an+1+an+2=(a1+a2+a3)*q^(n-1)>1/9关键是求q说实在的,我
2a^2(下标n+1)+3a(下标n+1)*an-2an^2=0[2a(n+1)-an]*[a(n+1)+an]=0因各项均为正数所以2a(n+1)=an即{an}是公比为1/2的等比数列又a3+1/
选择B.你的公式中,n-1是不是在a的下面.要是的话:a2=a1+2,得a2=3,依次a3=5,a4=7,.,所以通项公式为2n-1
sn=a1*(1-q^n)/(1-q)带入a1=1,q=a-3/2,sn=a(n无穷大)(1-(a-3/2)^n)/(5/2-a)=a因为(a-3/2)^n当n无穷大时存在,所以有-1
因为6Sn=(an+1)(an+2)(1)所以6Sn-1=(an-1+1)(an-1+2)(2)(1)-(2)则an-an-1=3所以an是等差数列因为6Sn=(an+1)(an+2)可知S1=a1=
(1)an,bn^2,an+1成等差数列2bn^2=an+a(n+1)bn^2,an+1,bn+1^2成等比数列a(n+1)^2=bn^2*b(n+1)^2a(n+1)=bnb(n+1)2bn^2=a
∵S50=9∴a1+a2+…+a50=9∵T50=107∴(a1+1)2+(a2+1)2+…+(a50+1)2=107即a12+a22+…+a502+2(a1+a2+…+a50)+50=107∴a12
a1+a2+...+an=(1/2)(an²+an)a1+a2+...+a(n-1)=(1/2)(a(n-1)²+a(n-1))两式相减得an=(1/2)(an²+an)
An=A1*q(n-1),An+1=A1*qn,An+2=A1*q(n+1),代入得q(n-1)=qn+q(n+1),消去的q2+q=1可解得q
证明:当n=1时S1=1/2(a1+1/a1)S1=a1所以a1=1/2(a1+1/a1)a1²=1因为a1>0所以左式=a1=1右式=√1-√1-1=1左式=右式所以假设n=k时,等式成立