已知数列(Xn),(Yn)满足X1=X2=1,Y1=Y2=2并且X(n+1)/Xn=已知数列{xn }、{yn}满足x1=x2=1 ,y1=y2=2并且x(n+1)/xn=λxn/x(n-1),y(n+1)/yn>=λyn/y(n-1)λ为非零实数,n=2,3,4,…).(1)若x1、x3、x5成等比数列,求参数λ的
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![已知数列(Xn),(Yn)满足X1=X2=1,Y1=Y2=2并且X(n+1)/Xn=已知数列{xn }、{yn}满足x1=x2=1 ,y1=y2=2并且x(n+1)/xn=λxn/x(n-1),y(n+1)/yn>=λyn/y(n-1)λ为非零实数,n=2,3,4,…).(1)若x1、x3、x5成等比数列,求参数λ的](/uploads/image/z/11509646-14-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BC%88Xn%EF%BC%89%2C%EF%BC%88Yn%EF%BC%89%E6%BB%A1%E8%B6%B3X1%3DX2%3D1%2CY1%3DY2%3D2%E5%B9%B6%E4%B8%94X%28n%2B1%29%2FXn%3D%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Bxn+%7D%E3%80%81%7Byn%7D%E6%BB%A1%E8%B6%B3x1%3Dx2%3D1+%2Cy1%3Dy2%3D2%E5%B9%B6%E4%B8%94x%28n%2B1%29%2Fxn%3D%CE%BBxn%2Fx%28n-1%29%2Cy%28n%2B1%29%2Fyn%3E%3D%CE%BByn%2Fy%28n-1%29%CE%BB%E4%B8%BA%E9%9D%9E%E9%9B%B6%E5%AE%9E%E6%95%B0%2Cn%3D2%2C3%2C4%2C%E2%80%A6%EF%BC%89%EF%BC%8E%EF%BC%881%EF%BC%89%E8%8B%A5x1%E3%80%81x3%E3%80%81x5%E6%88%90%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%2C%E6%B1%82%E5%8F%82%E6%95%B0%CE%BB%E7%9A%84)
已知数列(Xn),(Yn)满足X1=X2=1,Y1=Y2=2并且X(n+1)/Xn=已知数列{xn }、{yn}满足x1=x2=1 ,y1=y2=2并且x(n+1)/xn=λxn/x(n-1),y(n+1)/yn>=λyn/y(n-1)λ为非零实数,n=2,3,4,…).(1)若x1、x3、x5成等比数列,求参数λ的
已知数列(Xn),(Yn)满足X1=X2=1,Y1=Y2=2并且X(n+1)/Xn=
已知数列{xn }、{yn}满足x1=x2=1 ,y1=y2=2并且x(n+1)/xn=λxn/x(n-1),y(n+1)/yn>=λyn/y(n-1)λ为非零实数,n=2,3,4,…).
(1)若x1、x3、x5成等比数列,求参数λ的值;
(2)当λ>0时,证明:x(n+1)/y(n+1)<=xn/yn .(n∈N*)
(3)当λ>1时,证明(x1-y1)/(x2-y2)+(x2-y2)/(x3-y3)+省略号+(xn-yn)/(xn+1-yn+1)<λ/(λ-1)(n∈N*).
已知数列(Xn),(Yn)满足X1=X2=1,Y1=Y2=2并且X(n+1)/Xn=已知数列{xn }、{yn}满足x1=x2=1 ,y1=y2=2并且x(n+1)/xn=λxn/x(n-1),y(n+1)/yn>=λyn/y(n-1)λ为非零实数,n=2,3,4,…).(1)若x1、x3、x5成等比数列,求参数λ的
(1)x3=λ,x4=λ^3,x5=λ^6,
由x1,x3,x5成等比数列,得
λ^2=λ^6,λ≠0,
∴λ=土1.
(2)λ>0时易知xn,yn>0,
由x
由y
∴x
(3)?
x3=λ,x4=λ^3,x5=λ^6,
由x1,x3,x5成等比数列,得
λ^2=λ^6,λ≠0,
∴λ=土1.
(1)由已知x1=x2=1,且x3x2=λ
x2x1
∴x3=λ,同理可知x4=λ3,x5=λ6,若x1、x3、x5成等比数列,则x32=x1x5,即λ2=λ6.而λ≠0,解得λ=±1.
(2)证明:(Ⅰ)由已知λ>0,x1=x2=1及y1=y2=2,可得xn>0,yn>0.由不等式的性质,有yn+1yn≥λ
ynyn-1≥λ 2yn-1yn-2…≥
λ n...
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(1)由已知x1=x2=1,且x3x2=λ
x2x1
∴x3=λ,同理可知x4=λ3,x5=λ6,若x1、x3、x5成等比数列,则x32=x1x5,即λ2=λ6.而λ≠0,解得λ=±1.
(2)证明:(Ⅰ)由已知λ>0,x1=x2=1及y1=y2=2,可得xn>0,yn>0.由不等式的性质,有yn+1yn≥λ
ynyn-1≥λ 2yn-1yn-2…≥
λ n-1y2y1=λn-1;
另一方面,xn+1xn=λ
xnxn-1=λ 2xn-1xn-2…λ n-1x2x1=λn-1.
因此,yn+1yn≥λ n-1=xn+1xn(n∈N*).故xn+1yn+1≤
xnyn(n∈N*).
(Ⅱ)当λ>1时,由(Ⅰ)可知,yn>xn≥1(n∈N*).
又由(Ⅰ)xn+1yn+1≤
xnyn(n∈N*),则yn+1-xn+1xn+1≥yn-xnxn,
从而yn+1-xn+1yn-xn≥xn+1xn(n∈N*).
∴x1-y1x2-y2+
x2-y2x3-y3+…+
xn-ynxn+1-yn+1=
1-(
1λ)21-
1λ<
λλ-1(n∈N*)
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