Question

高二数列第二问,

Answer 1

(1)

L:{(x,y)| y=2x+1 }

Pn(an,bn)

x=0

y=1

P1=(0,1)

an = a1+(n-1)

= n-1

bn = 2an +1

= 2(n-1) +1

= 2n-1

(2)

|P1Pn|^2 = (bn -1 )^2 + (an)^2

= (2n-2)^2 + (n-1)^2

= 5(n-1)^2

|P1Pn| =√5(n-1)

cn =√5/[n.|P1Pn|]

= 1/[(n-1)n]

= 1/(n-1) -1/n

lim(n-> ∞) (c2+c3+...+cn)

=lim(n-> ∞) [ 1 - 1/n]

= 1