已知an=√1/(4n-3)?{bn}的前n项和为Tn?Tn+1/(an)^2=Tn/(an+1)^2+16n^2-8n-3?求当b1为多少时,{bn}为...已知an=√1/(4n-3)?{bn}的前n项和为Tn?Tn+1/(an)^2=Tn/(an+1)^2+16n^2-8n-3?求当b1为多少时,{bn}为等差数列

已知an=n,bn=1/3n,则数列{an/bn}的前n项和Sn= 已知{an},{bn}均为等差数列,前n项的和为An,Bn,且An/Bn=2n/(3n+1),求a10/b10的值 已知等差数列{an}和{bn},他们的前n项之和为An和Bn,若An/Bn=(5n+3)/(2n-1)A9/B9 已知两等差数列an.bn,且a1+a2+.+an/b1+b2+.+bn=3n+1/4n+3,对于任意正整数n都成立,求an：bn. 已知an=√1/(4n-3)?{bn}的前n项和为Tn?Tn+1/(an)^2=Tn/(an+1)^2+16n^2-8n-3?求当b1为多少时,{bn}为...已知an=√1/(4n-3)?{bn}的前n项和为Tn?Tn+1/(an)^2=Tn/(an+1)^2+16n^2-8n-3?求当b1为多少时,{bn}为等差数列 已知两个等差数列an和bn的前n项和分别为已知两个等差数列{an}和{bn}的前n项和分别是An和Bn,且An/Bn= (7n+45)/(n+3),则使得an/bn为整数的正整.为什么an/bn=An-1/Bn-1?这步不懂~ 已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2.求数列an的通向公式.设数列bn是的前n项和已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2.(1)求数列an的通向公式.(2)设数列bn是的前n项和为sn, 已知数列{an},{bn}满足:a1=3,当n>=2时,a(n-1)+an=4n;对于任意的正整数n,b1+2b2+…+2^(n-1)bn=nan.设{bn...已知数列{an},{bn}满足:a1=3,当n>=2时,a(n-1)+an=4n;对于任意的正整数n,b1+2b2+…+2^(n-1)bn=nan.设{bn}的前n项和为Sn 等差数列{an},{bn}的前n项和分别为An,Bn,切An/Bn=2n/3n+1,求lim(n→∞)an/bn 已知{an},{bn}是等差数列,他们的前N项和分别为An,Bn,An/Bn=2n/（3n+1),求lim an/bn的值 A2/3B1C根号6 /3D4/9 已知两个等差数列{an}和{bn}的前n项和分别为An和Bn,且An/Bn=(7n+1)/(4n+27）,则an/bn=? 已知各项均为正数的两个数列an,bn满足a n+1=an+bn/√an²+bn² 已知数列{an} {bn} {cn}满足(an+1-an)(bn+1-bn)=cn,n属于N*(1)设an=1/3^n,bn=1-3n,求数列{cn}的前n项和Sn(2)设cn=2n+4,{an}是公差为2的等差数列,若b1=1,求{bn}的通项公式(3)设cn=3n-25,an=n^2-8n,求正整数k使得对一切n属 已知数列{bn}前n项和Sn=3/2n^2-1/2n.数列{an}满足(an)^3=4^-(bn+2)(n ∈N*）,数列{cn}=anbn 求数列{cn}已知数列{bn}前n项和Sn=3/2n^2-1/2n.数列{an}满足(an)^3=4^-(bn+2)(n∈N*）,数列{cn}=anbn求数列an,bn通项公式和{cn}的 已知数列｛an｝和｛bn｝满足a1=m,an+1=λan+n,bn=an-2n/3+4/9已知数列｛an｝和｛bn｝满足a1=m,a（n+1)=λan+n,bn=an-2n/3+4/9（1）当m=1时,求证：于任意的实数λ,{an}一定不是等差数列（2）当λ=-1/2时,试判断{bn} 已知数列{an}是首项为a1=1/4,公比q=1/4的等比数列,设bn+2=3(log1/4)an(n∈N*),数列{Cn}满足Cn=an*bn求证：数列bn成等差数列 已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,求数列an的通向公式.已知数列{an}满足a1=-1,an=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2.求数列an的通向公式.第一遍打错了。是下面这个。an+1=[(3n+3)an+4n+6]/n,bn=3^(n-1)/an+2 已知数列{an}的前n项和Sn=n2+4n（n∈N*）,数列{bn}满足b1=1,bn+1=2bn+1（1）求数列{an},{bn}的通项公式；（2）设cn= (an-3)•(bn+1)4,求数列{cn}的前n项和Tn．