高中数学问题~~~

来源：百度知道 编辑：UC知道 时间：2024/06/28 11:54:34
设s，t为正整数，两条直线L1：t/(2s)*x+y-t=0与L2：t/(2s)*x-y=0的交点是（x1，y1），对于正整数n(n>=2),过点（0，t)和(Xn-1,0)的直线与直线L2的交点记为(Xn,Yn) 则xn-yn=？（用s，t，n表示） 要具体解答过程~~

Y(1) = t/2, X(1) = s.

过点（0，t)和(X(n-1),0)的直线为

(y - t)/t = x/X(n-1),

t/(2s)[(y-t)X(n-1)/t] - y = 0,

(y-t)X(n-1) - 2sy = 0,

Y(n) = tX(n-1)/[X(n-1) - 2s]

X(n) = X(n-1)(Y(n) - t)/t = X(n-1){tX(n-1)/[X(n-1) - 2s] - t}/t

= X(n-1){X(n-1)/[X(n-1) - 2s] - 1}

= 2sX(n-1)/[X(n-1) - 2s],

1/X(n) = [X(n-1) - 2s]/[2sX(n-1)] = 1/(2s) - 1/X(n-1),

1/X(2) = 1/(2s) - 1/X(1) = 1/(2s) - 1/s = -1/(2s)

1/X(3) = 1/(2s) - 1/X(2) = 1/(2s) + 1/(2s) = 1/s = 1/X(1),

1/X(4) = 1/(2s) - 1/X(3) = 1/(2s) - 1/X(1) = 1/X(2) = -1/(2s),

...
1/X(2n-1) = 1/X(1) = 1/s, X(2n-1) = s,
1/X(2n) = 1/X(2) = -1/(2s), X(2n) = -2s, n = 1,2,...

Y(2n-1) = tX(2n-2)/[X(2n-2) - 2s] = t(-2s)/[-4s] = t/2,

Y(2n) = tX(2n-1)/[X(2n-1) - 2s] = ts/[s - 2s] = -t.

X(2n-1) - Y(2n-1) = s - t/2,
X(2n) - Y(2n) = -2s + t,
n = 1,2,...

(Xn,Yn)过L2,L3&nbs