在直角坐标平面上有一系列p1(x1.y1),p2(x2,y2).Pn（Xn,Yn）对一切正整数n,点Pn位于函数y=3x

(1)
xn=-5/2+(n-1)(-1)=-n-3/2
yn=3xn+13/4=-3n-5/4
Pn为(-n-3/2,-3n-5/4)
(2)
设Cn方程为
y=a(x+(2n+3)/2)^2-(12n+5)/4
代入Dn(0,n^2+1)
a=1
y=x^2+(2n+3)x+n^2+1
y'=2x+2n+3
kn=y'(0)=2n+3
1/(k(n-1)*kn)=(1/2)*(1/(2n+1)-1/(2n+3))
1/k1*k2+1/k2*k3+…1/k(n-1)*kn=1/10-1/(4n+6)