设数列{an}和{bn}满足a1=b1=6,a2=b2=3,且数列{a(n+1)-an}是等差数列,数列{bn-2}是等比数列1 求数列{an}和{bn}的通项公式2 是否存在k属于正整数,使ak-bk属于（0,1/2）?若存在,求出k,若不存在,为什么?

设各项均为正数的数列{an}和{bn}满足:an,bn,an+1成等差数列,bn,an+1,bn+1等比数列且a1=1,b1=2,a2=3求通项an,bn 设各项均为正数的数列｛an}和｛bn}满足：an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列,且a1=1,b1=2,a2=3,求通项an,bn 设各项均为正数的数列{an}和{bn}满足5^[an ],5^[bn] ,5^[a(n+1)] .设各项均为正数的数列{an}和{bn}满足5^[an ],5^[bn] ,5^[a(n+1)] 成等比数列,lg[bn],lg[a(n+1)],lg[bn+1]成等差数列,且a1=1,b1=2,a2=3,求通项an、bn. 设数列{an}和{bn}满足:a1=b1=6,a2=b2=4,a3=b3=3,数列{an+1-an}是等差数列···设数列{an}和{bn}满足:a1=b1=6,a2=b2=4,a3=b3=3,数列{an+1-an}是等差数列,Sn为数列{bn}的前n项和,且Sn=2n-bn+10,(1)分别求{an}{bn}的通项公式(2 数列an,bn满足a1=b1=1,an+1-an=bn+1/bn=2,则数列ban的前10项和为 设数列an,bn分别满足a1*a2*a3...*an=1*2*3*4...*n,b1+b2+b3+...bn=an^2,n属于N+a1*a2*a3...*an=1*2*3*4...*n,b1+b2+b3+...bn=an^2,n属于N+1)求数列an和bn的通项公式 设数列an的前n项和为sn,且a1=1,an+1=2sn+1,数列bn满足a1=b1,点p(bn,bn+1)在直线x-y+2=0上,n是正整数.求an,bn的通项公式.设cn=bn/an,求cn的前n项和tn 设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3 ,且数列{an+1-an}是等差数列设数列{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3 ,且数列{a(n+1)-an}是等差数列,{bn-2}是等比数列(2)设{nbn}的前n项和为Sn,求Sn的表达式(3)数列{C 设数列An,Bn满足a1=b1=6,a2=b2=4,a3=b3=3,且数列A(n+1)-An(n属于正整数)是等差数列.设数列An,Bn满足a1=b1=6,a2=b2=4,a3=b3=3,且数列A(n+1)-An(n属于正整数)是等差数列,sn为数列｛BN}的前几项和,且sn=2n-bn+101）求数 设数列{an},{bn}满足a1=1,b1=0且(高二数学,a(n+1)=2an+3bn且b(n+1)=an+2bn.(1)求证:{an+根号3bn}和{an-根号3bn}都是等比数列并求其公比;(2)求{an},{bn}的通项公式(n均为正整数)是(根号3)bn 设数列{an}和{bn}都是等差数列,且a1=25,b1=75,a100+b100=100,则数列{an+bn}的前100项和为 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 已知数列an,bn满足a1=2/3,an+1=2an/an+2,b1+2b2+2^2b3++2^n-1bn=n(nN*) (1)求数列an和bn的通项公式； (2)设数列bn/an的前n项和Tn,问是否存在正整数m、M且M-m=3,使得m 设数列｛an｝和｛bn｝都是等差数列,且a1=25,b1=75,a2+b2=100,那么数列｛an+bn}的第37项为? 等差等比数列应用设数列{An}和{Bn}满足A1=B1=6,A2=B2=4,A3=B3=3,且数列{A（n+1）-An}是等差数列,数列{Bn-2}是等比数列（1）设,求数列{Cn}的通项公式（2）求数列{An}和{Bn}的通项公式 设数列{an}{bn}满足a1=b1=6 a2=b2=4 a3=b3=3若{an+1 - an}为等差数列.{bn+1 -bn}为等比数列.分别求{an}{bn}的通项公式. 已知数列{an}满足an+Sn=n,数列{bn}满足b1=a1,且bn=an-a(n-1),(n≥2）,试求数列{bn}的前n项的和Tn 设数列an为等比数列,数列bn满足bn=na1+(n-1)a2+...+2an-1+an已知b1=1,b2=4第一问为什么可以“由已知b1=a1”