设等比数列{an}的首项a1=256,前n项和为Sn,且Sn,Sn+2,Sn+1成等差数列.(I)求{an}的公比q (2)用iin表示{an}的前n项之积，即 IIn=a1*a2......an,试比较II7 II8 II9的大小 己知函数f(x)=x+t/x(t>0)和点P(1,0)，过点P作曲

设等比数列{an}的首项a1=256,前n项和为Sn,且Sn,Sn+2,Sn+1成等差数列.(I)求{an}的公比q (2)用iin表示{an}的前n项之积，即 IIn=a1*a2......an,试比较II7 II8 II9的大小 己知函数f(x)=x+t/x(t>0)和点P(1,0)，过点P作曲
设等比数列{an}的首项a1=256,前n项和为Sn,且Sn,Sn+2,Sn+1成等差数列.(I)求{an}的公比q (2)用iin表示{an}
的前n项之积，即
IIn=a1*a2......an,试比较II7 II8 II9的大小
己知函数f(x)=x+t/x(t>0)和点P(1,0)，过点P作曲线y=f(x)的两条切线PM、PN，切点分别为M、N
（1）设/MN/=g(t)试求函数g(t)的表达式；
(2)是否存在t，使得M、N与A(1,0)三点共线，若存在，求出t的值；若不存在，请说明理由
『3』在（1）的条件下，若对任意的正整数n，在区间[2,n+64/n]内总存在m+1个实数a1,a2,...,am,am+1,使得不等式g(a1)+g(a2)+...+g(am)

设等比数列{an}的首项a1=256,前n项和为Sn,且Sn,Sn+2,Sn+1成等差数列.(I)求{an}的公比q (2)用iin表示{an}的前n项之积，即 IIn=a1*a2......an,试比较II7 II8 II9的大小 己知函数f(x)=x+t/x(t>0)和点P(1,0)，过点P作曲
2S(n+2)=Sn+S(n+1)
2[Sn+a(n+1)+a(n+2)]=Sn+Sn+a(n+1)
2Sn+2a(n+1)+2a(n+2)=2Sn+a(n+1)
2a(n+1)+2a(n+2)=a(n+1)
a(n+1)+2a(n+2)=0
2a(n+2)=-a(n+1)
a(n+2)/a(n+1)=-1/2
即q=-1/2
an=a1q^(n-1)
=256*(-1/2)^(n-1)
=2^8*(-1/2)^(n-1)
=(-1/2)^-8*(-1/2)^(n-1)
=(-1/2)^(n-9)
=(-2)^(9-n)
IIn=a1*a2.an
II7=a1*a2.a7
=(-2)^8*(-2)^7*.*(-2)^2
=(-2)^[(2+8)*7/2]
=(-2)^35
II8=a1*a2.a8
=(-2)^8*(-2)^7*.*(-2)^1
=(-2)^[(1+8)*8/2]
=(-2)^36
II9=a1*a2.a9
=(-2)^8*(-2)^7*.*(-2)^0
=(-2)^[(0+8)*9/2]
=(-2)^36
II8=II9>II7
(1)
f(x)=x+t/x
f'(x)=1-t/x^2
设过A点与f(x)相切直线切点为(a,b)
切线斜率k=b/(a-1)=f'(a)=1-t/a^2
又f(a)=a+t/a=b
(a+t/a)/(a-1)=1-t/a^2
a^2+2at-t=0
t>0方程有两根,即为分别M,N横坐标,设为x1,x1
则M,N纵坐标分别为y1=x1+t/x1,y2=x2+t/x2
y1+y2=x1+x2+t/x1+t/x2=(x1+x2)+t(x1+x2)/x1x2
=-2t+t(-2t)/(-t)=0
y1y2=(x1+t/x1)(x2+t/x2)=x1x2+tx1/x2+tx2/x1+t^2/(x1x2)
=-t+t[(x1+x2)^2-2x1x2]/(-t)+t^2/(-t)
=-4t^2-4t
|MN|^2=(x1-x2)^2+(y1-y2)^2
=(x1+x2)^2-4x1x2+(y1+y2)^2-4y1y2
=(-2t)^2+4t+0+4(4t^2+4t)
=20(t^2+t)
g(t)=|MN|=2√[5(t^2+t)]
(2)
沿用(1)的设定和结论,M,N为不同两点x1≠x2
假设M,N,A共线,则直线AM,AN斜率相等
f'(x1)=f'(x2)
1-t/x1^2=1-t/x2^2
x1^2=x2^2
x1=-x2
又:
(x1+t/x1-1)/x1=(x2+t/x2-1)/x2
x1+t/x1-1=x1+t/x1+1
无解
所以不存在t使M,N,A共线
(3)
应当是在(1)的条件下
易知g(t)=2√[5(t^2+t)]在区间(2,n+64/n]单调递增
使不等式g(a1)+g(a2)+.g(am)=16,而对于任意正整数n都要成立
所以必须满足g(a1)+g(a2)+.g(am)mg(2)
所以:mg(2)

为m（x乘AM）

2S(n+2)=Sn+S(n+1)
2[Sn+a(n+1)+a(n+2)]=Sn+Sn+a(n+1)
2Sn+2a(n+1)+2a(n+2)=2Sn+a(n+1)
2a(n+1)+2a(n+2)=a(n+1)
a(n+1)+2a(n+2)=0
2a(n+2)=-a(n+1)
a(n+2)/a(n+1)=-1/2
即q=-1/2
an=...

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2S(n+2)=Sn+S(n+1)
2[Sn+a(n+1)+a(n+2)]=Sn+Sn+a(n+1)
2Sn+2a(n+1)+2a(n+2)=2Sn+a(n+1)
2a(n+1)+2a(n+2)=a(n+1)
a(n+1)+2a(n+2)=0
2a(n+2)=-a(n+1)
a(n+2)/a(n+1)=-1/2
即q=-1/2
an=a1q^(n-1)
=256*(-1/2)^(n-1)
=2^8*(-1/2)^(n-1)
=(-1/2)^-8*(-1/2)^(n-1)
=(-1/2)^(n-9)
=(-2)^(9-n)
IIn=a1*a2......an
II7=a1*a2......a7
=(-2)^8*(-2)^7*....*(-2)^2
=(-2)^[(2+8)*7/2]
=(-2)^35
II8=a1*a2......a8
=(-2)^8*(-2)^7*....*(-2)^1
=(-2)^[(1+8)*8/2]
=(-2)^36
II9=a1*a2......a9
=(-2)^8*(-2)^7*....*(-2)^0
=(-2)^[(0+8)*9/2]
=(-2)^36
II8=II9>II7

(1)
f(x)=x+t/x
f'(x)=1-t/x^2
设过A点与f(x)相切直线切点为(a,b)
切线斜率k=b/(a-1)=f'(a)=1-t/a^2
又f(a)=a+t/a=b
(a+t/a)/(a-1)=1-t/a^2
a^2+2at-t=0
t>0方程有两根,即为分别M,N横坐标,设为x1,x1
则M,N纵坐标分别为y1=x1+t/x1,y2=x2+t/x2
y1+y2=x1+x2+t/x1+t/x2=(x1+x2)+t(x1+x2)/x1x2
=-2t+t(-2t)/(-t)=0
y1y2=(x1+t/x1)(x2+t/x2)=x1x2+tx1/x2+tx2/x1+t^2/(x1x2)
=-t+t[(x1+x2)^2-2x1x2]/(-t)+t^2/(-t)
=-4t^2-4t
|MN|^2=(x1-x2)^2+(y1-y2)^2
=(x1+x2)^2-4x1x2+(y1+y2)^2-4y1y2
=(-2t)^2+4t+0+4(4t^2+4t)
=20(t^2+t)
g(t)=|MN|=2√[5(t^2+t)]
(2)
沿用(1)的设定和结论,M,N为不同两点x1≠x2
假设M,N,A共线,则直线AM,AN斜率相等
f'(x1)=f'(x2)
1-t/x1^2=1-t/x2^2
x1^2=x2^2
x1=-x2
又:
(x1+t/x1-1)/x1=(x2+t/x2-1)/x2
x1+t/x1-1=x1+t/x1+1
无解
所以不存在t使M,N,A共线
(3)
应当是在(1)的条件下
易知g(t)=2√[5(t^2+t)]在区间(2,n+64/n]单调递增
使不等式g(a1)+g(a2)+......g(am)m要最大则a(i)尽量小,a(m+1)尽量大
又n+64/n>=16,而对于任意正整数n都要成立
所以必须满足g(a1)+g(a2)+......g(am)又g(a1)+g(a2)+......g(am)>mg(2)
所以:mg(2)m2√30<2√1360
m<√(136/3),m<=6

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