设a1=2,a2=4,数列{bn}满足:bn=an+1-an,bn+1=2bn+2.
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/24 11:42:19
设a1=2,a2=4,数列{bn}满足:bn=an+1-an,bn+1=2bn+2.
(1)求b1、b2;
(2)求证数列{bn+2}是等比数列(要指出首项与公比);
(3)求数列{an}的通项公式.
(1)求b1、b2;
(2)求证数列{bn+2}是等比数列(要指出首项与公比);
(3)求数列{an}的通项公式.
(1)b1=a2-a1=4-2=2,b2=2b1+2=2×2+2=6.
(2)证明:∵bn+1=2bn+2,∴bn+2+2=2(bn+2).
∴数列{bn+2}是以b1+2=4为首项,2为公比的等比数列.
(3)由(2)可得:bn+2=4×2n−1=2n+1.
∴bn=2n+1−2.
∴an−an−1=2n−2.
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=(2n-2)+(2n-1-2)+…+(22-2)+2
=2n+2n-1+…+22+2-2(n-1)
=
2(2n−1)
2−1-2n+2
=2n+1-2n.
(2)证明:∵bn+1=2bn+2,∴bn+2+2=2(bn+2).
∴数列{bn+2}是以b1+2=4为首项,2为公比的等比数列.
(3)由(2)可得:bn+2=4×2n−1=2n+1.
∴bn=2n+1−2.
∴an−an−1=2n−2.
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=(2n-2)+(2n-1-2)+…+(22-2)+2
=2n+2n-1+…+22+2-2(n-1)
=
2(2n−1)
2−1-2n+2
=2n+1-2n.
急 设A1=2,A2=4,数列Bn满足:Bn=A(n+1)-An,B(n+1)=2Bn +2
急 设A1=2,A2=4,数列BN满足:Bn=A(n+1)-An,B(n+1)=2Bn+2
设a1=2,a2=4,数列{bn}满足:bn=a(n+1)-an,b(n+1)=2bn+2.
设A1=2,A2=4,数列{Bn}满足:Bn=A(n+1) –An,B(n+1)=2Bn+2.
数列{an}和{bn}满足a1=1 a2=2 an>0 bn=根号an*an+1
一道数学数列题设两个数列{An},{Bn}满足Bn=(A1+A2+A3+……+nAn)/(1+2+3+……+),若{Bn
已知数列{an}、{bn}满足:a1=1/4,an+bn=1,bn+1=bn/1-an^2 (1)求{an}的通项公式
数列{an}和{bn}满足a1=1 a2=2 an>0 bn=根号an*an+1且{bn}是以公比为q的等比数列
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
设数列an,bn分别满足a1*a2*a3...*an=1*2*3*4...*n,b1+b2+b3+...bn=an^2,
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)
已知数列{an}和{bn}满足:a1=1,a2=2,an>0,bn=根号anan+1,且{bn}是以q为公比的等比数列.