已知点pn(an,bn)满足a n+1=anb n+1,b n+1=bn/1-4an^2,且p1的坐标为(1,-1).(1)求过点P1 P2的直线L解析式(2)已知pn(an,bn)在L上,求对于所有n∈N*,能使不等式(1+a1)(1+a2)...(1+an)>=k*根号下(1/b2b3...bn+1)成立的k
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![已知点pn(an,bn)满足a n+1=anb n+1,b n+1=bn/1-4an^2,且p1的坐标为(1,-1).(1)求过点P1 P2的直线L解析式(2)已知pn(an,bn)在L上,求对于所有n∈N*,能使不等式(1+a1)(1+a2)...(1+an)>=k*根号下(1/b2b3...bn+1)成立的k](/uploads/image/z/10440608-32-8.jpg?t=%E5%B7%B2%E7%9F%A5%E7%82%B9pn%28an%2Cbn%29%E6%BB%A1%E8%B6%B3a+n%2B1%3Danb+n%2B1%2Cb+n%2B1%3Dbn%2F1-4an%5E2%2C%E4%B8%94p1%E7%9A%84%E5%9D%90%E6%A0%87%E4%B8%BA%EF%BC%881%2C-1%EF%BC%89.%EF%BC%881%EF%BC%89%E6%B1%82%E8%BF%87%E7%82%B9P1+P2%E7%9A%84%E7%9B%B4%E7%BA%BFL%E8%A7%A3%E6%9E%90%E5%BC%8F%EF%BC%882%EF%BC%89%E5%B7%B2%E7%9F%A5pn%28an%2Cbn%29%E5%9C%A8L%E4%B8%8A%2C%E6%B1%82%E5%AF%B9%E4%BA%8E%E6%89%80%E6%9C%89n%E2%88%88N%2A%2C%E8%83%BD%E4%BD%BF%E4%B8%8D%E7%AD%89%E5%BC%8F%281%2Ba1%29%281%2Ba2%29...%281%2Ban%29%3E%3Dk%2A%E6%A0%B9%E5%8F%B7%E4%B8%8B%281%2Fb2b3...bn%2B1%29%E6%88%90%E7%AB%8B%E7%9A%84k)
已知点pn(an,bn)满足a n+1=anb n+1,b n+1=bn/1-4an^2,且p1的坐标为(1,-1).(1)求过点P1 P2的直线L解析式(2)已知pn(an,bn)在L上,求对于所有n∈N*,能使不等式(1+a1)(1+a2)...(1+an)>=k*根号下(1/b2b3...bn+1)成立的k
已知点pn(an,bn)满足a n+1=anb n+1,b n+1=bn/1-4an^2,且p1的坐标为(1,-1).
(1)求过点P1 P2的直线L解析式
(2)已知pn(an,bn)在L上,求对于所有n∈N*,能使不等式
(1+a1)(1+a2)...(1+an)>=k*根号下(1/b2b3...bn+1)成立的k的最大值
已知点pn(an,bn)满足a n+1=anb n+1,b n+1=bn/1-4an^2,且p1的坐标为(1,-1).(1)求过点P1 P2的直线L解析式(2)已知pn(an,bn)在L上,求对于所有n∈N*,能使不等式(1+a1)(1+a2)...(1+an)>=k*根号下(1/b2b3...bn+1)成立的k
解析:(1)Pn(an,an+1)(n∈N*)在一次函数y=2x+k的图象上,an+1=2an+k,即an+1+k=2(an+k),
又bn=an+1-an=an+k,则bn+1=an+1+k,
所以==2,故数列{bn}是等比数列.
(2)由(1),b1=a1+k,bn=b1×2n-1=(a1+k)2n-1,an=bn-k,
则S6=T6-6k=-6k=63a1+57k,
T4==15(a1+k),
由S6=T4得a1=-k.
又S5=-9,即T5-5k=-9,-5k=-9,
即31a1+26k=-9.
将a1=-k代入,得k=8.