设数列An,Bn 满足a1=b1=6,a2=b2=4,a3=b3=3设数列An,Bn 满足a1=b1=6,a2=b2=4,a3=b3=3,若{an+1 -an}是等差数列,{bn+1-bn}是等比数列.1.求An Bn 通项2.求数列AN最小项及最小值3.是否存在K属于N*,使ak-bk属于（0.0.5）若存

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}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. 数列{an}{bn}满足an=5an-1 -6bn-1 bn=3an-1 -4bn-1 且a1=a,b1=b求{an}{bn}通项 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. 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}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 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? 设数列An,Bn 满足a1=b1=6,a2=b2=4,a3=b3=3设数列An,Bn 满足a1=b1=6,a2=b2=4,a3=b3=3,若{an+1 -an}是等差数列,{bn+1-bn}是等比数列.1.求An Bn 通项2.求数列AN最小项及最小值3.是否存在K属于N*,使ak-bk属于（0.0.5）若存 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 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}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)分别求{an}{bn}的通项公式(2)是否存在k∈N*,使bk-ak∈(0,1/2)?若存在,求出k；若不存在,说明理由. 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” 已知数列｛an｝中,a1=1,a(n+1)=2Sn+1(n属于N*),数列｛bn｝中,b1=1,b4=7,且满足b(n+2)+bn=2b(n+1),求数列｛an｝与｛bn｝的通项公式.设Sn是数列｛an·bn｝的前n项和,求Sn 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