设{an}和{bn}满足a1=b1=6,a2=b2=4,a3=b3=3
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设{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,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N使Ak-Bk属于(0,1/2)
且数列〔an+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N使Ak-Bk属于(0,1/2)
1)
a2-a1=4-6=-2
a3-a2=3-4=-1
(a3-a2)-(a2-a1)=-1-(-2)=1
〔an+1-an〕是公差为1的等差数列
an+1-an==-2+(n-1)=n-3
an-a(n-1)=(n-1)-3=n-4
an=a(n-1)+(n-4)
=a(n-2)+(n-5)+(n-4)
=...
=a1+(-2)+(-1)+...+(n-5)+(n-4)
=a1+(n-1)(n-6)/2
=6+(n^2-7n+6)/2
=(n^2-7n+18)/2
b1-2=4
b2-2=2
b3-2=1
数列{bn-2}是公比为1/2的等比数列
bn-2=(b1-2)*(1/2)^(n-1)=4*(1/2)^(n-1)=(1/2)^(n-3)
bn=2+(1/2)^(n-3)
2)
Ak-Bk=(k^2-7k+18)/2-2-(1/2)^(k-3)
=(k^2-7k+14)/2-(1/2)^(k-3)
a2-a1=4-6=-2
a3-a2=3-4=-1
(a3-a2)-(a2-a1)=-1-(-2)=1
〔an+1-an〕是公差为1的等差数列
an+1-an==-2+(n-1)=n-3
an-a(n-1)=(n-1)-3=n-4
an=a(n-1)+(n-4)
=a(n-2)+(n-5)+(n-4)
=...
=a1+(-2)+(-1)+...+(n-5)+(n-4)
=a1+(n-1)(n-6)/2
=6+(n^2-7n+6)/2
=(n^2-7n+18)/2
b1-2=4
b2-2=2
b3-2=1
数列{bn-2}是公比为1/2的等比数列
bn-2=(b1-2)*(1/2)^(n-1)=4*(1/2)^(n-1)=(1/2)^(n-3)
bn=2+(1/2)^(n-3)
2)
Ak-Bk=(k^2-7k+18)/2-2-(1/2)^(k-3)
=(k^2-7k+14)/2-(1/2)^(k-3)
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