已知正项数列{an},{bn}满足:a1=3,a2=6,{bn}是等差数列,且对任意正整数n,都有bn,根号an,bn+
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/04/30 00:34:14
已知正项数列{an},{bn}满足:a1=3,a2=6,{bn}是等差数列,且对任意正整数n,都有bn,根号an,bn+1成等比数
列.(1)求数列{bn}的通项公式.(2)设Sn=1/a1+1/a2+.+1/an,试比较Sn与1的大小.求哪位大仙给解下,
列.(1)求数列{bn}的通项公式.(2)设Sn=1/a1+1/a2+.+1/an,试比较Sn与1的大小.求哪位大仙给解下,
(1)bn,√an,bn+1成等比
所以an=bn*bn+1
所以a1=b1*b2=3 a2=b2*b3=6
所以b1*(b1+d)=3 (b1+d)*(b1+2d)=6
解得:b1=√2 d=√2/2或者b1=-√2 d=-√2/2舍去
所以bn=b1+(n-1)d=√2(n+1)/2
(2)an=bn*bn+1=(n+1)(n+2)/2
所以1/an=2/(n+1)(n+2)=2(1/(n+1)-1/(n+2))
所以Sn=1/a1+1/a2+.+1/an
=2(1/2-1/3+1/3-1/4……+1/(n+1)-1/(n+2))
=1-2/(n+2)
所以an=bn*bn+1
所以a1=b1*b2=3 a2=b2*b3=6
所以b1*(b1+d)=3 (b1+d)*(b1+2d)=6
解得:b1=√2 d=√2/2或者b1=-√2 d=-√2/2舍去
所以bn=b1+(n-1)d=√2(n+1)/2
(2)an=bn*bn+1=(n+1)(n+2)/2
所以1/an=2/(n+1)(n+2)=2(1/(n+1)-1/(n+2))
所以Sn=1/a1+1/a2+.+1/an
=2(1/2-1/3+1/3-1/4……+1/(n+1)-1/(n+2))
=1-2/(n+2)
已知正项数列{an},{bn}满足:a1=3,a2=6,{bn}是等差数列,且对任意正整数n,都有bn,根号an,bn+
已知正项数列{an}{bn}满足,对任意正整数n,都有an,bn,an+1成等差数列,bn,an+1,bn+1成等比数列
已知正项数列{an},{bn}满足:对任意正整数n,都有an,bn,a(n+1)成等差数列,bn,a(n+1),b(n+
有两个正数数列an,bn,对任意正整数n,有an,bn,an+1成等比数列,bn,an+1,bn+1成等差数列,若a1=
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a
已知数列{an}是等差数列,且a1=2,a1+a2+a3=12 令bn=an*3^n,求{bn}的前n项和
已知数列{an}{bn}满足a1=1,a2=3,b(n+1)/bn=2,bn=a(n+1)-an,(n∈正整数),求数列
已知数列{an}和{bn}满足关系:bn=(a1+a2+a3+…+an)/n,(n∈N*).若{bn}是等差数列,求证{
已知等比数列{an}的通项公式为an=3^(n-1),设数列{bn}满足对任意自然数n都有b1/a1+b2/a2+b3/
已知数列an,bn满足a1=1,a2=3,(b(n)+1)/bn=2,bn=a(n+1)-an,(n∈正整数)
设数列{An}{Bn} 满足A1=B1= A2=B2=6 A3=B3=5且{An+1-An}是等差数列{Bn+1-Bn}
设数列{an}、{bn}满足:a1=b1=6,a2=b2=4,a3=b3=3,且数列{an+1-an}是等差数列,{bn