M(-3,0)N(3,0)B(1,0) 动圆C与MN切于B 过M ,N与C相切的两直线交于P 求P轨迹等差数列a2=5,a6=21,1/a的前n项和为Sn 若S(2n+1)-Sn≤m/15 则正整数m最小值为?
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![M(-3,0)N(3,0)B(1,0) 动圆C与MN切于B 过M ,N与C相切的两直线交于P 求P轨迹等差数列a2=5,a6=21,1/a的前n项和为Sn 若S(2n+1)-Sn≤m/15 则正整数m最小值为?](/uploads/image/z/1804220-44-0.jpg?t=M%EF%BC%88-3%2C0%EF%BC%89N%EF%BC%883%2C0%29B%281%2C0%29+%E5%8A%A8%E5%9C%86C%E4%B8%8EMN%E5%88%87%E4%BA%8EB+%E8%BF%87M+%2CN%E4%B8%8EC%E7%9B%B8%E5%88%87%E7%9A%84%E4%B8%A4%E7%9B%B4%E7%BA%BF%E4%BA%A4%E4%BA%8EP+%E6%B1%82P%E8%BD%A8%E8%BF%B9%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97a2%3D5%2Ca6%3D21%2C1%2Fa%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn+%E8%8B%A5S%EF%BC%882n%2B1%EF%BC%89-Sn%E2%89%A4m%2F15+%E5%88%99%E6%AD%A3%E6%95%B4%E6%95%B0m%E6%9C%80%E5%B0%8F%E5%80%BC%E4%B8%BA%3F)
M(-3,0)N(3,0)B(1,0) 动圆C与MN切于B 过M ,N与C相切的两直线交于P 求P轨迹等差数列a2=5,a6=21,1/a的前n项和为Sn 若S(2n+1)-Sn≤m/15 则正整数m最小值为?
M(-3,0)N(3,0)B(1,0) 动圆C与MN切于B 过M ,N与C相切的两直线交于P 求P轨迹
等差数列a2=5,a6=21,1/a的前n项和为Sn 若S(2n+1)-Sn≤m/15 则正整数m最小值为?
M(-3,0)N(3,0)B(1,0) 动圆C与MN切于B 过M ,N与C相切的两直线交于P 求P轨迹等差数列a2=5,a6=21,1/a的前n项和为Sn 若S(2n+1)-Sn≤m/15 则正整数m最小值为?
由已知,设PM,PN分别与圆C相切于R、Q,根据圆的切线长定理,有PQ=PR,MQ=MB,NR=NB;
所以
PM-PN=QM-RN=MB-NB=21)
设等差数列{an}公差为d.
a6-a2=4d=21-5=16 d=4
a1=a2-d=5-4=1
an=a1+(n-1)d=1+4(n-1)=4n-3
1/an=1/(4n-3)
S(2n+1)-Sn=a1+a2+...+an+a(n+1)+...+a(2n+1)-(a1+a2+...+an)
=a(n+1)+a(n+2)+...+a(2n+1)
a[(n+1)+1]+a[(n+1)+2]+...+a[2(n+1)+1]-[a(n+1)+a(n+2)+...+a(2n+1)]
=a(n+2)+a(n+3)+...+a(2n+1)+a(2n+2)+a(2n+3)-[a(n+1)+a(n+2)+...+a(2n+1)]
=a(2n+2)+a(2n+3)-a(n+1)
=1/[4(2n+2) -3]+1/[4(2n+3) -3] -1/[4(n+1)-3]
=1/(8n+5)+1/(8n+9)-2/(8n+2)