已知数列{xn},{yn}满足x1=y1=1,x2=y2=2,并且xn+1-(λ+1)xn+λxn-1=0,yn+1-(
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已知数列{xn},{yn}满足x1=y1=1,x2=y2=2,并且xn+1-(λ+1)xn+λxn-1=0,yn+1-(λ+1)yn+λyn-1≥0(�
已知数列{xn},{yn}满足x1=y1=1,x2=y2=2,并且xn+1-(λ+1)xn+λxn-1=0,yn+1-(λ+1)yn+λyn-1≥0(λ为非零参数,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xn+1-yn+1≤xn-yn(n∈N*);
(3)设0<λ<1,k∈N*,证明:(x2-x1)+(x4-x2)+(x6-x3)+…+(x2k-xk)<
已知数列{xn},{yn}满足x1=y1=1,x2=y2=2,并且xn+1-(λ+1)xn+λxn-1=0,yn+1-(λ+1)yn+λyn-1≥0(λ为非零参数,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xn+1-yn+1≤xn-yn(n∈N*);
(3)设0<λ<1,k∈N*,证明:(x2-x1)+(x4-x2)+(x6-x3)+…+(x2k-xk)<
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(1?λ)
(1)∵x1=1,x2=2,xn+1-(λ+1)xn+λxn-1=0,
∴x3=(λ+1)x2-λx1=2(λ+1)-λ=λ+2. x4=(λ+1)(λ+2)-2λ=λ2+λ+2. x5=(λ+1)(λ2+λ+2)?λ(λ+2)=λ3+λ2+λ+2. ∵x1,x3,x5成等比数列,∴ x23=x1?x5, ∴(λ+2)2=1×(λ2+λ+2),解得λ=-2. (2)下面利用数学归纳法证明: ①当n=1时,x2-x1=2-1=y2-y1,∴x2-y2≤x1-y1成立; ②假设当n=k时,xk+1-xk≤yk+1-yk成立,即xk+1-yk+1≤xk-yk成立. 则当n=k+1时,∵λ>0,∴xk+2-xk+1=λ(xk+1-xk)≤λ(yk+1-yk)≤yk+2-yk+1成立, 即xk+2-yk+2≤xk+1-yk+1成立. 即命题定义n=k+1时也成立. 综上可知:命题定义任意n∈N*都成立. (3)由xn+1-(λ+1)xn+λxn-1=0,可得xn+1-xn=λ(xn-xn-1), ∴xn?xn?1=(x2?x1)?λn?2=λn?2, ∴xn=(xn-xn-1)+(xn-1-xn-2)+…+(x3-x2)+(x2-x1)+x1 =λn-2+λn-3+…+λ+1+1= λn?1?1 λ?1+1.(0<λ<1). ∴x2k= λ2k?1?1 λ?1+1. ∴x2k-xk= λ2k?1?λk λ?1 ∴左边=(x2-x1)+(x4-x2)+(x6-x3)+…+(x2k-xk) = 1 λ?1[(λ+λ3+…+λ2k-1)-(λ+λ2+…+λk)] = 1 λ?1[ λ(λ2k?1) λ2?1? λk?1 λ?1] = 1 1?λ (1?λk)(1?λk+1) 1?λ2 = 1 (1?λ)2? (1?λk)(1?λk+1) 1+λ< 1 (1?λ)2.=右边. 故不等式成立.
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