Question

高二数学求解~~~~~！！求高手进

答案是123都对

Answer 1

P0(x0,y0)
x0>0,y0 >0
P(n-1)(x(n-1),y(n-1))
Since P(n-1) is on y= x^2
y(n-1) = (x(n-1))^2 (1)

y= x^2
y' = 2x
y'(x(n-1)) = 2x(n-1)
equation of tangent at (x(n-1), y(n-1) )
y- y(n-1) = 2x(n-1) . [ x- x(n-1) ]
y=0, x =xn
y(n-1) = -2x(n-1) .( xn-x(n-1)) (2)
from (1),(2)
-2x(n-1) .( xn-x(n-1)) = (x(n-1))^2
-2x(n-1)xn = -[x(n-1)]^2
xn/x(n-1) = 1/2
xn/x0 = (1/2)^n
xn = x0. (1/2)^n

Since Pn is on y=x^2
yn = (xn)^2
= (x0)^2 . (1/2)^(2n)
(1) xn > 0 ： 对
(2)
xn 是单调递减数列 ： 对
(3)
y0+y1+y2+..+yn
=(1/3)(x0)^2(1 - (1/2)^2n)
lim(n-> ∞) y0+y1+y2+..+yn = (1/3)x0^2
choose 1<x0 < √6
y0+y1+y2+..+yn <2
对