作业帮设数列{an}和{bn}满足:a1=b1=6,a2=b2=4,a3=b3=3,数列{an+1-an}是等差数列···
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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)是否存在k∈N*,使ak-bk∈(0,1/2)?若存在,求出k；若不存在,说明理由.
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)是否存在k∈N*,使ak-bk∈(0,1/2)?若存在,求出k；若不存在,说明理由.
（1）数列{an+1-an}是等差数列
a2-a1=4-6=-2 a3-a2=3-4=-1
公差=(a3-a2)-(a2-a1)=-1+2=1
则a(n+1)-an=-2+(n-1)*1=n-3
an-a(n-1)=n-4
.
a3-a2=3-4=-1
a2-a1=2-4=-2
叠加an-a1=(-2+n-4)*(n-1)/2=(1/2)n^2-7n/2+3
通项an=(1/2)(n^2-7n+18)
Sn=2n-bn+10 S(n-1)=2n-2-b(n-1)+10
bn=-bn+b(n-1)+2 2bn=b(n-1)+2
2(bn-2)=b(n-1)-2
{bn-2}是公比为(1/2)等比数列
则bn-2=(b1-2)*(1/2)^(n-1)=4*(1/2)^(n-1)
通项bn=(1/2)^(n-3)+2
（2） 设存在k∈N*,使ak-bk∈(0,1/2)
则ak-bk=(1/2)(k^2-7k+18)-(1/2)^(k-3)-2
=(1/2)(k^2-7k+14)-(1/2)^(k-3)
=(1/2)[(k^2-7k+14)-(1/2)^(k-2)]
设f(x)=k^2-7k+14=(k-7/2)^2+7/4
为开口向上的抛物线,最小值在顶点处f(7/2)=7/4
由于k取自然数,则f(3)=f(4)=2为最小
设g(x)=(1/2)^(k-2)
g'(x)=-(1/2)^(k-1)ln2
a2-a1=4-6=-2 a3-a2=3-4=-1
公差=(a3-a2)-(a2-a1)=-1+2=1
则a(n+1)-an=-2+(n-1)*1=n-3
an-a(n-1)=n-4
.
a3-a2=3-4=-1
a2-a1=2-4=-2
叠加an-a1=(-2+n-4)*(n-1)/2=(1/2)n^2-7n/2+3
通项an=(1/2)(n^2-7n+18)
Sn=2n-bn+10 S(n-1)=2n-2-b(n-1)+10
bn=-bn+b(n-1)+2 2bn=b(n-1)+2
2(bn-2)=b(n-1)-2
{bn-2}是公比为(1/2)等比数列
则bn-2=(b1-2)*(1/2)^(n-1)=4*(1/2)^(n-1)
通项bn=(1/2)^(n-3)+2
（2） 设存在k∈N*,使ak-bk∈(0,1/2)
则ak-bk=(1/2)(k^2-7k+18)-(1/2)^(k-3)-2
=(1/2)(k^2-7k+14)-(1/2)^(k-3)
=(1/2)[(k^2-7k+14)-(1/2)^(k-2)]
设f(x)=k^2-7k+14=(k-7/2)^2+7/4
为开口向上的抛物线,最小值在顶点处f(7/2)=7/4
由于k取自然数,则f(3)=f(4)=2为最小
设g(x)=(1/2)^(k-2)
g'(x)=-(1/2)^(k-1)ln2
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}、{bn}满足：a1=b1=6，a2=b2=4，a3=b3=3，且数列{an+1-an}是等差数列，{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···
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数列an是等差数列,bn是等比数列,满足b1=a1^2,b2=a2^2,b3=a3^2,求数列bn公比q